case study of introduction to trigonometry class 10

Class 10 Maths Case Study Questions Chapter 8 Introduction to Trigonometry

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Case study Questions in the Class 10 Mathematics Chapter 8  are very important to solve for your exam. Class 10 Maths Chapter 8 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 Maths Case Study Questions Chapter 8  Introduction to Trigonometry

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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Introduction to Trigonometry Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

Case Study/Passage-Based Questions

Question 1:

Answer: (d) 6m

(ii) Measure of ∠A =

Answer: (c) 45°

(iii) Measure of ∠C =

(iv) Find the value of sinA + cosC.

Answer: (d) 2√2

(v) Find the value of tan 2 C + tan 2  A.

Answer: (c) 2

Question 2:

Answer: (a) 30°

(ii) The measure of  ∠C is

Answer: (c) 60°

(iii) The length of AC is 

Answer: (d)6√3m

(iv) cos2A =

Answer: (b)1/2

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Introduction to Trigonometry Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Case Study on Introduction to Trigonometry Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Introduction to Trigonometry Class 10 Maths can use this page to download the PDF file. 

The case study questions on Introduction to Trigonometry are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Introduction to Trigonometry case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Introduction to Trigonometry Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Introduction to Trigonometry, therefore, they prepared a set of solutions along with the case study questions on Introduction to Trigonometry.

The case study on Introduction to Trigonometry Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Introduction to Trigonometry case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Introduction to Trigonometry Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Introduction to Trigonometry case study questions on Class 10 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Introduction to Trigonometry Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Introduction to Trigonometry case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Introduction to Trigonometry Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Introduction to Trigonometry as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Introduction to Trigonometry?

Students can choose their own way to answer Case Study on Introduction to Trigonometry Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Introduction to Trigonometry Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Introduction to Trigonometry questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Introduction to Trigonometry Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Introduction to Trigonometry?

 A few essential things to know to solve Case Study Questions on Class 10 Introduction to Trigonometry are -

• Basic Formulas of Introduction to Trigonometry: One of the most important things to know to solve Case Study Questions on Class 10 Introduction to Trigonometry is to learn about the basic formulas or revise them before solving the case-based questions on Introduction to Trigonometry.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Introduction to Trigonometry case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Introduction to Trigonometry Class 10 Maths?

Use Selfstudys.com to find Case Study on Introduction to Trigonometry Class 10 Maths. For ease, here is a step-wise procedure to download the Introduction to Trigonometry Case Study for Class 10 Maths in PDF for free of cost.

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Case Study and Passage Based Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry

• Last modified on: 9 months ago
• Reading Time: 3 Minutes

Case Study Questions:

Question 1:

Ananya is feeling so hungry and so thought to eat something. She looked into the fridge and found a bread pieces. She decided to make a sandwich. She cut the piece of bread diagonally and found it forms a right-angled triangle, with sides 4 cm, 4√3 cm and 8 cm.

On the basis of above information, answer the following questions.

(i) The value of ∠M is

D. None of these

(ii) The value of ∠K is

(iii) Find the value of tan M.

(iv) sec 2 M – 1 =

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Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables C hapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

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CBSE Class 10 Maths Case Study Questions for Chapter 9 - Some Applications of Trigonometry (Published By CBSE)

Check case study questions for cbse class 10 maths chapter 9 - some applications of trigonometry. these questions are published by the cbse itself for class 10 students..

Case study based questions are new for class 10 students. Therefore, it is quite essential that students practice with more of such questions so that they do not have problem in solving them in their Maths board exam. We have provided here the case study questions for CBSE Class 10 Maths Chapter 9 - Some Applications of Trigonometry. All these questions have been published by the Central Board of Secondary Education (CBSE) for the class 10 students. Therefore, students must solve all the questions seriously so that they may score the desired marks in their Maths exam.

Check Case Study Questions for Class 10 Maths Chapter 9:

CASE STUDY 1:

A group of students of class X visited India Gate on an education trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kingsway), is about 138 feet (42 metres) in height.

1. What is the angle of elevation if they are standing at a distance of 42m away from the monument?

Answer: b) 45 o

2. They want to see the tower at an angle of 60 o . So, they want to know the distance where they should stand and hence find the distance.

Answer: a) 25.24 m

3. If the altitude of the Sun is at 60 o , then the height of the vertical tower that will cast a shadow of length 20 m is

a) 20√3 m

b) 20/ √3 m

c) 15/ √3 m

d) 15√3 m

Answer: a) 20√3 m

4. The ratio of the length of a rod and its shadow is 1:1. The angle of elevation of the Sun is

5. The angle formed by the line of sight with the horizontal when the object viewed is below the horizontal level is

a) corresponding angle

b) angle of elevation

c) angle of depression

d) complete angle

Answer: a) corresponding angle

CASE STUDY 2:

A Satellite flying at height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka, them being Nanda Devi(height 7,816m) and Mullayanagiri (height 1,930 m). The angles of depression from the satellite, to the top of Nanda Devi and Mullayanagiri are 30° and 60° respectively. If the distance between the peaks of the two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains.

1. The distance of the satellite from the top of Nanda Devi is

a) 1139.4 km

b) 577.52 km

d) 1025.36 km

Answer: a) 1139.4 km

2. The distance of the satellite from the top of Mullayanagiri is

Answer: c) 1937 km

3. The distance of the satellite from the ground is

Answer: b) 577.52 km

4. What is the angle of elevation if a man is standing at a distance of 7816m from Nanda Devi?

5.If a mile stone very far away from, makes 45 o to the top of Mullanyangiri mountain. So, find the distance of this mile stone from the mountain.

a) 1118.327 km

b) 566.976 km

Also Check:

Case Study Questions for All Chapters of CBSE Class 10 Maths

Tips to Solve Case Study Based Questions Accurately

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CBSE Class 10 Maths: Case Study Questions of Chapter 8 Introduction to Trigonometry PDF Download

Case study Questions in the Class 10 Mathematics Chapter 8  are very important to solve for your exam. Class 10 Maths Chapter 8 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based   questions for Class 10 Maths Chapter 8  Introduction to Trigonometry

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Introduction to Trigonometry Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

Case Study/Passage-Based Questions

Question 1:

Answer: (d) 6m

(ii) Measure of ∠A =

Answer: (c) 45°

(iii) Measure of ∠C =

(iv) Find the value of sinA + cosC.

Answer: (d) 2√2

(v) Find the value of tan 2 C + tan 2  A.

Answer: (c) 2

Question 2:

Answer: (a) 30°

(ii) The measure of  ∠C is

Answer: (c) 60°

(iii) The length of AC is 

Answer: (d)6√3m

(iv) cos2A =

Answer: (b)1/2

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Introduction to Trigonometry Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Chapter 8 Class 10 Introduction to Trignometry

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Trigonometry means studying relationship between measures of triangle. Usually, we talk about right triangles when we study trigonometry,

In this chapter, we will study

• What is sin, cos, tan ( Sine, cosine, tangent) ... and how they are found in a triangle
• What is sec, cosec, cot, and how is it related to sin, cos, tan.
• (Sin, cos, tan, sec, cosec, cot are known as Trigonometric Ratios)
• Then, we study Trigonometric ratios of specific angles l ike 0°, 30°, 45°, 60°, 90° ; and do some questions
• We study the formulas of sin (90 - θ) , cos (90 - θ), tan (90 - θ)
• And then we study Trigonometric Identities, and how other identities are derived from sin 2 θ + cos 2 θ = 1

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Class 10 Maths Chapter 8 Case Based Questions - Introduction to Trigonometry

Case study - 1.

Based on the above information give the, answer of the following questions:

Case Study - 2

This is a single repeating triangle in a truss system.

Case Study - 3

Q5: Find the value of tan 2 C + tan 2 A. (a) 0 (b) 1 (c) 2 (d) 1/2 Ans:  (c) Explanation:  tanC = 1, tanA = tan45 ∘ = 1 ⇒ tan 2 C + tan 2 A = 1 + 1 = 2.

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Class 10th Maths - Introduction to Trigonometry Case Study Questions and Answers 2022 - 2023

Class 10th Maths - Introduction to Trigonometry Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Introduction to Trigonometry, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Introduction to trigonometry case study questions with answer key.

Final Semester - June 2015

(ii) The measure of   \(\angle\) C is

(iii) The length of AC is 

(iv) cos2A =

(v) sin  \(\left(\frac{C}{2}\right)\) =

(ii) Find cot B

(iii) Find tanA.

(iv) Find secA.

(v) Find cosecB.

(ii) The value of sec  \(\theta\)   =

(iii) The value of  \(\frac{\tan \theta}{1+\tan ^{2} \theta}=\)

(iv) The value of  \(\cot ^{2} \theta-\operatorname{cosec}^{2} \theta=\)  

(v) The value of  \(\sin ^{2} \theta+\cos ^{2} \theta=\)

(ii) The value of \(\angle\) K =

(iii) Find the value of tanM.

(iv) sec 2 M - 1 =

(v) The value of  \(\frac{\tan ^{2} 45^{\circ}-1}{\tan ^{2} 45^{\circ}+1}\)   is

(ii) Measure of \(\angle\) A =

(iii) Measure of \(\angle\) C =

(iv) Find the value of sinA + cosC.

(v) Find the value of tan 2 C + tan 2 A.

*****************************************

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Class 10 Case study Chapter 8 introduction to Trigonometry

Case Study:1

An electrician wanted to repair a street lamp at a height of 15 feet. He places his ladder such that its foot is 8 feet from the foot of the lamp post as shown in the figure below:

(A) Find the value of ‘cos R’.

(B) Find the value of ‘cosec P’.

(A) In ΔPQR ∠Q = 90°

We will first calculate PR by using pythagoras theorem in ΔPQR

(B) Again from In ΔPQR

Case Study :2

Built in 13th century the magnificent Qutub-Minar in Delhi, in Red and buff sandstone is the highest tower in India. It is an architectural marval of Ancient India. Qutub-ud-Din Aibak of Slave Dynasty laid the foundation of Minar in A.D. 1199 for the use of Mu’azzin(crier) to give calls for prayer and raised the first storey, to which were added three more storeys by his successor and son-in-law. Shams-uud-Din Itutmish (A.D. 1211-36).

Let us take the height PQ of Qutub Minar as 72 m for ease of calculations (though actual height is 72.5 m) and distance of point r from Q as 65 m.

(A) The value of ‘cos R’ is:

(B) The value of ‘cot P’ is:

(a) 1                                  (b) -1

(c) 0                                 (d) 2

Explanation:

In Right angle triangle PQR using Pythagoras theorem

(B) Answer (c) 72/65

From (A) PR = 97 m, PQ = 72 m, QR = 65 m

( C) Answer (a) 1

Explanation: 

(D) Answer(c) 333/20

Case study :3

Varanasi is a city of temples, including the gold-plated Vishwanath sacred to Shiva; the Bharat Mata, or Mother India, temple that boasts a huge three-dimensional relief map of the Indian subcontinent carved out of marble; and the hundreds of small temples that dot the waterways and alleys. It is a city of school, home to one of Asia’s largest universities. It is also a city of legends.

The figure below shows one such temple along the banks of the sacred river “Ganges” or “Ganga”. A person sitting at point marked A looks at the top of nearby temple and imagines that a right angled triangle ABC can be drawn as shown in the figure below.

Let the distance between the person and the base of the temple be 12 m and the height of the temple be 5 m.

(a) 12/13             (b) 13/12

(c) 5/13               (d) 13/5

(a) 12/13              (b) 5/13

(c) 13/12               (d) 13/5

(A) Answer (c) 5/13

Explanation: We know,

In ΔABC Using Pythagoras theorem

(B) We have,

(C) Answer (a) 12/13

Some other Case study question:

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Class 10 Case based problem of Chapter 5 A.P. 1

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CBSE Case Study Questions for Class 10 Maths Trigonometry Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Trigonometry  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

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CBSE Case Study Questions for Class 10 Maths Trigonometry PDF

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• NCERT Solutions for Class 10 Maths Chapter 8- Introduction To Trigonometry
• NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Free PDF Download

NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry is provided here on Vedantu. Chapter 8 consists of the discussion of basic trigonometry, opposite & adjacent sides in a right-angled triangle, basic trigonometric ratios, and standard values of trigonometric ratios and complementary trigonometric ratios. Students can download the Class 10 Maths Chapter 8 NCERT Solutions PDF for free from Vedantu.

Here, we have provided the important formulas and key concepts to help students revise the chapter before attempting NCERT Solutions. Also, the NCERT solutions are curated by experts and have easy step-wise solutions to the questions.

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Exercises under NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 "Introduction to Trigonometry" contains four exercises. Below is a brief explanation of each exercise:

Exercise 8.1 - In this exercise, students are introduced to the basic trigonometric ratios - sine, cosine, and tangent, and their reciprocal functions. They learn how to find the values of these ratios for acute angles in a right triangle. The exercise also covers the concept of trigonometric identities, which are fundamental relationships between trigonometric functions.

Exercise 8.2 - This exercise deals with the application of trigonometric ratios to solve real-life problems. Students learn how to use trigonometric ratios to find the height and distance of an object, as well as the angle of elevation and depression. They also learn how to use the Pythagorean theorem to solve problems involving right triangles.

Exercise 8.3 - In this exercise, students learn about complementary angles and how to find the trigonometric ratios of complementary angles. They also learn about the trigonometric ratios of angles greater than 90 degrees and less than 0 degrees, and how to find their values using the reference angle.

Exercise 8.4 - The final exercise covers the concept of trigonometric equations. Students learn how to solve trigonometric equations using the identities and ratios learned in the previous exercises. They also learn how to find the general solution of a trigonometric equation, which involves finding all the possible solutions. Finally, the exercise covers the concept of the period of a trigonometric function and how to find it for different functions.

Access NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry

Exercise 8.1 

1. In $\Delta ABC$ right angled at $B$, $AB=24\text{ cm}$, $BC=7\text{ cm}$. Determine 

(i) $\sin A,\cos A$ 

Ans: Given that in the right angle triangle $\Delta ABC$, $AB=24\text{ cm}$, $BC=7\text{ cm}$.

Let us draw a right triangle $\Delta ABC$, also $AB=24\text{ cm}$, $BC=7\text{ cm}$. We get

We have to find $\sin A,\cos A$.

We know that for right triangle

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and 

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

Here, $AB=24\text{ cm}$, $BC=7\text{ cm}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( 24 \right)}^{2}}+{{\left( 7 \right)}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=576+49$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=625\text{ }c{{m}^{2}}$

$\Rightarrow AC=25\text{ cm}$ 

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ 

$\Rightarrow \sin A=\dfrac{BC}{AC}$ 

$\therefore \sin A=\dfrac{7}{25}$

$\Rightarrow \cos A=\dfrac{AB}{AC}$

$\therefore \cos A=\dfrac{24}{25}$

(ii) $\sin C,\cos C$ 

We have to find $\sin C,\cos C$.

$\Rightarrow AC=25\text{ cm}$

$\Rightarrow \sin C=\dfrac{AB}{AC}$ 

$\therefore \sin C=\dfrac{24}{25}$

$\Rightarrow \cos C=\dfrac{BC}{AC}$

$\therefore \cos A=\dfrac{7}{25}$

3. In the given figure find $\tan P-\cot R$.

Ans: Given in the figure, 

$PQ=12\text{ cm}$ 

$PQ=13\text{ cm}$

$\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$ and 

$\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$

Now, we need to apply the Pythagoras theorem to find the measure of adjacent side/base.

In $\Delta PQR$, by Pythagoras theorem ,

$\Rightarrow {{\left( PR \right)}^{2}}={{\left( PQ \right)}^{2}}+{{\left( QR \right)}^{2}}$ 

$\Rightarrow {{\left( 13 \right)}^{2}}={{\left( 12 \right)}^{2}}+{{\left( QR \right)}^{2}}$

$\Rightarrow 169=144+{{\left( QR \right)}^{2}}$

$\Rightarrow {{\left( QR \right)}^{2}}=169-144$

$\Rightarrow {{\left( QR \right)}^{2}}=25\text{ }c{{m}^{2}}$

$\Rightarrow QR=5\text{ cm}$ 

$\tan P=\dfrac{\text{opposite side}}{\text{adjacent side}}$ 

$\Rightarrow \tan P=\dfrac{QR}{PQ}$ 

$\therefore \tan P=\dfrac{5}{12}$

$\cot R=\dfrac{\text{adjacent side}}{\text{opposite side}}$

$\Rightarrow \cot R=\dfrac{QR}{PQ}$

$\therefore \cot R=\dfrac{5}{12}$

$\Rightarrow \tan P-\cot R=\dfrac{5}{12}-\dfrac{5}{12}$

$\therefore \tan P-\cot R=0$ 

3. If $\sin A=\dfrac{3}{4}$, calculate $\cos A$ and $\tan A$.

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

Given that $\sin A=\dfrac{3}{4}$.

We know that  $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$.

From the above figure, we get

$\sin A=\dfrac{BC}{AC}$

Therefore, we get

$\Rightarrow BC=3$ and

$\Rightarrow AC=4$ 

Now, we have to find the values of $\cos A$ and $\tan A$.

We know that $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ and $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$.

Here, $AC=4\text{ cm}$, $BC=3\text{ cm}$

$\Rightarrow {{4}^{2}}=A{{B}^{2}}+{{3}^{2}}$

$\Rightarrow 16=A{{B}^{2}}+9$

$\Rightarrow A{{B}^{2}}=16-9$

$\Rightarrow A{{B}^{2}}=7$

$\Rightarrow AB=\sqrt{7}\text{ cm}$

Now, we get

$\cos A=\dfrac{AB}{AC}$ 

$\therefore \cos A=\dfrac{\sqrt{7}}{4}$ 

And $\tan A=\dfrac{BC}{AB}$

$\therefore \tan A=\dfrac{3}{\sqrt{7}}$

4. Given $15\cot A=8$. Find $\sin A$ and $\sec A$.

Given that $15\cot A=8$.

We get $\cot A=\dfrac{8}{15}$.

We know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$.

$\cot A=\dfrac{AB}{BC}$

$\Rightarrow BC=15$ and

$\Rightarrow AB=8$ 

Now, we have to find the values of $\sin A$ and $\sec A$.

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\sec \theta =\dfrac{\text{hypotenuse}}{\text{adjacent side}}$.

$\Rightarrow A{{C}^{2}}={{8}^{2}}+{{15}^{2}}$

$\Rightarrow A{{C}^{2}}=64+225$

$\Rightarrow A{{C}^{2}}=289$

$\Rightarrow AC=17\text{ cm}$

$\sin A=\dfrac{BC}{AC}$ 

$\therefore \sin A=\dfrac{15}{17}$ 

And $\sec A=\dfrac{AC}{AB}$

$\therefore \sec A=\dfrac{17}{8}$

5. Given $\sec \theta =\dfrac{13}{12}$, calculate all other trigonometric ratios.

Given that $\sec \theta =\dfrac{13}{12}$.

We know that $\sec \theta =\dfrac{\text{hypotenuse}}{\text{adjacent side}}$.

$\sec \theta =\dfrac{AC}{AB}$

$\Rightarrow AC=13$ and

$\Rightarrow AB=12$ 

Now, we need to apply the Pythagoras theorem to find the measure of the perpendicular/opposite side.

$\Rightarrow {{13}^{2}}={{12}^{2}}+B{{C}^{2}}$

$\Rightarrow 169=144+B{{C}^{2}}$

$\Rightarrow B{{C}^{2}}=25$

$\Rightarrow BC=5\text{ cm}$

Now, we know that

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$

Here, $\sin \theta =\dfrac{BC}{AC}$ 

$\therefore \sin \theta =\dfrac{5}{13}$ 

We know that $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Here, $\cos \theta =\dfrac{AB}{AC}$ 

$\therefore \cos \theta =\dfrac{12}{13}$

We know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, $\tan \theta =\dfrac{BC}{AB}$ 

$\therefore \tan \theta =\dfrac{5}{12}$

We know that $\operatorname{cosec}\theta =\dfrac{\text{hypotenuse}}{\text{opposite side}}$

Here, $\operatorname{cosec}\theta =\dfrac{AC}{BC}$

$\therefore \operatorname{cosec}\theta =\dfrac{13}{5}$

We know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$

Here, $\cot \theta =\dfrac{\text{AB}}{BC}$

\[\therefore \cot \theta =\dfrac{12}{5}\] .

6. If $\angle A$ and $\angle B$ are acute angles such that $\cos A=\cos B$, then show that $\angle A=\angle B$.

Given that $\cos A=\cos B$.

In a right triangle $\Delta ABC$, we know that 

Here, 

$\cos A=\dfrac{AC}{AB}$ 

And $\cos B=\dfrac{BC}{AB}$ 

As given $\cos A=\cos B$, we get

$\Rightarrow \dfrac{AC}{AB}=\dfrac{BC}{AB}$ 

$\Rightarrow AC=AB$ 

Now, we know that angles opposite to the equal sides are also equal in measure.

Then, we get

$\angle A=\angle B$ 

Hence proved.

7. Evaluate the following if $\cot \theta =\dfrac{7}{8}$

(i) $\dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}$ 

Now, in a right triangle we know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$.

Here, from the figure $\cot \theta =\dfrac{BC}{AB}$ .

$AB=8$ and 

$BC=7$ 

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{8}^{2}}+{{7}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=64+49$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=113$

$\Rightarrow AC=\sqrt{113}$ 

Now, we know that 

Here, we get

$\sin \theta =\dfrac{AB}{AC}=\dfrac{8}{\sqrt{113}}$ and 

$\cos \theta =\dfrac{BC}{AC}=\dfrac{7}{\sqrt{113}}$ 

Now, we have to evaluate 

$\dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}$

Applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }\] 

Substituting the values, we get

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{1-{{\left( \dfrac{8}{\sqrt{113}} \right)}^{2}}}{1-{{\left( \dfrac{7}{\sqrt{113}} \right)}^{2}}}\]

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{1-\dfrac{64}{113}}{1-\dfrac{49}{113}}\]

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{\dfrac{113-64}{113}}{\dfrac{113-49}{113}}\]

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\]

\[\therefore \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{49}{64}\]

(ii) ${{\cot }^{2}}\theta $ 

Ans: Given that $\cot \theta =\dfrac{7}{8}$.

Now, ${{\cot }^{2}}\theta ={{\left( \dfrac{7}{8} \right)}^{2}}$

$\therefore {{\cot }^{2}}\theta =\dfrac{49}{64}$ 

8. If $3\cot A=4$, check whether $\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$ or not.

Given that $3\cot A=4$.

We get $\cot A=\dfrac{4}{3}$.

$\Rightarrow AB=4$ 

$\Rightarrow A{{C}^{2}}={{4}^{2}}+{{3}^{2}}$

$\Rightarrow A{{C}^{2}}=16+9$

$\Rightarrow A{{C}^{2}}=25$

$\Rightarrow AC=5$

Now, let us consider LHS of the expression $\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$, we get

$LHS=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}$

Now, we know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

$\tan A=\dfrac{BC}{AB}=\dfrac{3}{4}$ 

Substitute the value, we get

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{1-{{\left( \dfrac{3}{4} \right)}^{2}}}{1+{{\left( \dfrac{3}{4} \right)}^{2}}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{\dfrac{16-9}{16}}{\dfrac{16+9}{16}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{\dfrac{7}{16}}{\dfrac{25}{16}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{7}{25}$

Now, let us consider RHS of the expression $\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$, we get

$RHS={{\cos }^{2}}A-{{\sin }^{2}}A$

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$.

$\sin A=\dfrac{BC}{AC}=\dfrac{3}{5}$  

And $\cos A=\dfrac{AB}{AC}=\dfrac{4}{5}$

Substitute the values, we get

$\Rightarrow {{\cos }^{2}}A-{{\sin }^{2}}A={{\left( \dfrac{4}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}$

$\Rightarrow {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{16}{25}-\dfrac{9}{25}$

$\Rightarrow {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{7}{25}$

Hence, we get LHS=RHS

$\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$.

9. In $ABC$, right angled at $B$. If $\tan A=\dfrac{1}{\sqrt{3}}$, find the value of 

(i) $\sin A\cos C+\cos A\sin C$ 

Given that $\tan A=\dfrac{1}{\sqrt{3}}$.

In a right triangle, we know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, from the figure we get

$\tan A=\dfrac{BC}{AB}=\dfrac{1}{\sqrt{3}}$

We get $BC=1$ and $AB=\sqrt{3}$ .

$\Rightarrow A{{C}^{2}}={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}$

$\Rightarrow A{{C}^{2}}=3+1$

$\Rightarrow A{{C}^{2}}=4$

$\Rightarrow AC=2$

$\sin A=\dfrac{BC}{AC}=\dfrac{1}{2}$  and $\sin C=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$

And $\cos A=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$and $\cos C=\dfrac{BC}{AC}=\dfrac{1}{2}$

Now, we have to find the value of the expression $\sin A\cos C+\cos A\sin C$.

Substituting the values we get

$\Rightarrow \sin A\cos C+\cos A\sin C=\dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}$ 

$\Rightarrow \sin A\cos C+\cos A\sin C=\dfrac{1}{4}+\dfrac{3}{4}$

$\Rightarrow \sin A\cos C+\cos A\sin C=\dfrac{4}{4}$

$\therefore \sin A\cos C+\cos A\sin C=1$

(ii) $\cos A\cos C-\sin A\sin C$

Now, we have to find the value of the expression $\cos A\cos C-\sin A\sin C$.

$\Rightarrow \cos A\cos C-\sin A\sin C=\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}$ 

$\Rightarrow \cos A\cos C-\sin A\sin C=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}$

$\therefore \Rightarrow \cos A\cos C-\sin A\sin C=0$

10. In $\Delta PQR$, right angled at $Q$, $PR+QR=25\text{ cm}$ and $PQ=5\text{ cm}$. Determine the values of $\sin P,\cos P$ and $\tan P$.

Ans: Let us consider a right angled triangle $\Delta PQR$, we get

Given that $PR+QR=25\text{ cm}$ and $PQ=5\text{ cm}$.

Let $QR=25-PR$

Now, applying the Pythagoras theorem in $\Delta PQR$, we get

$\Rightarrow P{{R}^{2}}={{5}^{2}}+{{\left( 25-PR \right)}^{2}}$

$\Rightarrow P{{R}^{2}}=25+{{25}^{2}}+P{{R}^{2}}-50PR$

$\Rightarrow P{{R}^{2}}=P{{R}^{2}}+25+625-50PR$

$\Rightarrow 50PR=650$

$\Rightarrow PR=13\text{ cm}$ 

Therefore, 

$\Rightarrow QR=12\text{ cm}$ 

Now, we know that in right triangle,

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$, $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ and $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$.

$\sin P=\dfrac{QR}{PR}$ 

$\therefore \sin P=\dfrac{12}{13}$ 

$\cos P=\dfrac{PQ}{PR}$ 

$\therefore \cos P=\dfrac{5}{13}$ 

$\tan P=\dfrac{QR}{PQ}$ 

$\therefore \tan P=\dfrac{12}{5}$ 

11. State whether the following are true or false. Justify your answer.

(i) The value of $\tan A$ is always less than $1$.

Ans: The given statement is false. The value of $\tan A$ depends on the length of sides of a right triangle and sides of a triangle may have any measure.

(ii) For some value of angle $A$, $\sec A=\dfrac{12}{5}$.

Ans: We know that in the right triangle $\sec A=\dfrac{\text{hypotenuse}}{\text{adjacent side of }\angle \text{A}}$ .

We know that in the right triangle the hypotenuse is the largest side.

Therefore, the value of $\sec A$ must be greater than $1$.

In the given statement $\sec A=\dfrac{12}{5}$, which is greater than $1$.

Therefore, the given statement is true.

(iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.

Ans: The given statement is false because $\cos A$ is the abbreviation used for the cosine of angle $A$. Abbreviation used for the cosecant of angle $A$ is $\operatorname{cosec}A$.

(iv) $\cot A$ is the product of $\cot $ and $A$.

Ans: $\cot A$ is the abbreviation used for the cotangent of angle $A$. Hence the given statement is false.

(v) For some angle $\theta $, $\sin \theta =\dfrac{4}{3}$.

Ans: We know that in the right triangle $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ .

Therefore, the value of $\sin \theta $ must be less than $1$.

In the given statement $\sin \theta =\dfrac{4}{3}$, which is greater than $1$.

Therefore, the given statement is false.

Exercise 8.2

1. Evaluate the following:

(i) $\sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ $ 

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $\sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ $.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}$

$\Rightarrow \dfrac{3}{4}+\dfrac{1}{4}$ 

$\Rightarrow \dfrac{4}{4}$ 

$\therefore \sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ =1$.

(ii) $2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ $

We have to evaluate $2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ $.

$\Rightarrow 2{{\left( 1 \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}$

$\Rightarrow 2+\dfrac{3}{4}-\dfrac{3}{4}$ 

$\Rightarrow 2$ 

$\therefore 2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ =2$.

(iii) $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$

We have to evaluate $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$.

$\Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}$

$\Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}}$ 

\[\Rightarrow \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\]

Multiplying and dividing by \[\sqrt{3}-1\], we get

\[\Rightarrow \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}\]

\[\Rightarrow \dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{2}\left( 2+2\sqrt{3} \right)\left( \sqrt{3}-1 \right)}\]

\[\Rightarrow \dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{2\sqrt{2}\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)}\]

\[\Rightarrow \dfrac{3-\sqrt{3}}{2\sqrt{2}\left( {{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}} \right)}\]

\[\Rightarrow \dfrac{3-\sqrt{3}}{2\sqrt{2}\left( 3-1 \right)}\]

$\Rightarrow \dfrac{3-\sqrt{3}}{4\sqrt{2}}$ 

$\therefore \dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }=\dfrac{3-\sqrt{3}}{4\sqrt{2}}$

(iv) $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }$

We have to evaluate $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }$.

$\Rightarrow \dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1}$

$\Rightarrow \dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{3}{2}}$

$\Rightarrow \dfrac{\dfrac{3\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{3\sqrt{3}+4}{2\sqrt{3}}}$

$\Rightarrow \dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}$

Multiplying and dividing by \[3\sqrt{3}-4\], we get

\[\Rightarrow \dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \dfrac{3\sqrt{3}-4}{3\sqrt{3}-4}\]

Now, applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \dfrac{{{\left( 3\sqrt{3}-4 \right)}^{2}}}{{{\left( 3\sqrt{3} \right)}^{2}}-{{4}^{2}}}\]

\[\Rightarrow \dfrac{27+16-24\sqrt{3}}{27-16}\]

$\Rightarrow \dfrac{43-24\sqrt{3}}{11}$ 

$\therefore \dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }=\dfrac{43-24\sqrt{3}}{11}$

(v) $\dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }$

We have to evaluate $\dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }$.

$\Rightarrow \dfrac{5{{\left( \dfrac{1}{2} \right)}^{2}}+4{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}-{{1}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}$

$\Rightarrow \dfrac{5\left( \dfrac{1}{4} \right)+4\left( \dfrac{4}{3} \right)-1}{\left( \dfrac{1}{4} \right)+\left( \dfrac{3}{4} \right)}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{1+3}{4}}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{4}{4}}$ 

$\Rightarrow \dfrac{\dfrac{67}{12}}{1}$ 

$\therefore \dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }=\dfrac{67}{12}$.

2. Choose the correct option and justify your choice.

(i) $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=$ ………

(a) $\sin 60{}^\circ $ 

(b) $\cos 60{}^\circ $ 

(c) $\tan 60{}^\circ $ 

(d) $\sin 30{}^\circ $ 

Ans: The given expression is $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Substitute the value in the given expression we get

$\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\sqrt{3}}{2}$

From the trigonometric table we know that 

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ 

$\cos 60{}^\circ =\dfrac{1}{2}$ 

$\tan 60{}^\circ =\sqrt{3}$ 

$\sin 30{}^\circ =\dfrac{1}{2}$

Hence, $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\sin 60{}^\circ $.

Therefore, option (A) is the correct answer.

(ii) $\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=$ ………

(a) $\tan 90{}^\circ $ 

(b) $1$ 

(c) $\sin 45{}^\circ $ 

Ans: The given expression is $\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 45{}^\circ =1$.

$\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{1-{{1}^{2}}}{1+{{1}^{2}}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{1-1}{1+1}$

$\Rightarrow \Rightarrow \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{0}{2}$

$\therefore \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=0$

Therefore, option (D) is the correct answer.

(iii) $\sin 2A=2\sin A$ is true when $A=$ ……..

(a) $0{}^\circ $ 

(b) $30{}^\circ $ 

(c) $45{}^\circ $ 

(d) $60{}^\circ $ 

Ans: The given expression is $\sin 2A=2\sin A$.

We know that from the trigonometric ratio table we have 

$\sin 0{}^\circ =0$

$\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}$

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$  

$\sin 90{}^\circ =1$ 

The given statement is true when $A=0{}^\circ $.

$\Rightarrow \sin 2A=2\sin A$

$\Rightarrow \sin 2\times 0{}^\circ =2\sin 0{}^\circ $

$0=0$ 

(iv) $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=$………

Ans: The given expression is $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }$.

$\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\sqrt{3}$

Hence, $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\tan 60{}^\circ $.

Therefore, option (C) is the correct answer.

3. If $\tan \left( A+B \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$, $0{}^\circ <A+B\le 90{}^\circ $. Find $A$ and $B$ .

Ans: Given that $\tan \left( A+ dB \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$.

From the trigonometric ratio table we know that $\tan 60{}^\circ =\sqrt{3}$ and $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Then we get

$\tan \left( A+B \right)=\sqrt{3}$

$\Rightarrow \tan \left( A+B \right)=\tan 60{}^\circ $

$\Rightarrow A+B=60{}^\circ $ ……….(1)

Also, $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \tan \left( A-B \right)=\tan 30{}^\circ $

$\Rightarrow A-B=30{}^\circ $ ……….(2)

Adding eq. (1) and (2), we get

$2A=90{}^\circ $

$\therefore A=45{}^\circ $ 

Substitute the obtained value in eq. (1), we get

$45{}^\circ +B=60{}^\circ $ 

$\Rightarrow B=60{}^\circ -45{}^\circ $ 

$\therefore B=15{}^\circ $ 

Therefore, the values of $A$ and $B$ is $45{}^\circ $ and $15{}^\circ $ respectively.

4. State whether the following are true or false. Justify your answer.

(i) $\sin \left( A+B \right)=\sin A+\sin B$.

Ans: Let us assume $A=30{}^\circ $ and $B=60{}^\circ $.

Now, let us consider LHS of the given expression, we get

$\sin \left( A+B \right)$

Substitute the assumed values in the LHS, we get

$\sin \left( A+B \right)=\sin \left( 30{}^\circ +60{}^\circ  \right)$

$\Rightarrow \sin \left( A+B \right)=\sin \left( 90{}^\circ  \right)$ 

From the trigonometric ratio table we know that $\sin 90{}^\circ =1$, we get

$\Rightarrow \sin \left( A+B \right)=1$

Now, let us consider the RHS of the given expression and substitute the values, we get

$\sin A+\sin B=\sin 30{}^\circ +\sin 60{}^\circ $

From the trigonometric ratio table we know that $\sin 30{}^\circ =\dfrac{1}{2}$ and $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$, we get

$\Rightarrow \sin A+\sin B=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}$

$\Rightarrow \sin A+\sin B=\dfrac{1+\sqrt{3}}{2}$

Thus, $LHS\ne RHS$.

(ii) The value of $\sin \theta $ increases as $\theta $ increases. 

Ans: The value of sine from the trigonometric ratio table is as follows:

$\sin 30{}^\circ =\dfrac{1}{2}=0.5$

$\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}=0.707$

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}=0.866$  

Therefore, we can conclude that the value of $\sin \theta $ increases as $\theta $ increases. 

(iii) The value of $\cos \theta $ increases as $\theta $ increases. 

Ans: The value of cosine from the trigonometric ratio table is as follows:

$\cos 0{}^\circ =1$

$\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}=0.866$

$\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}=0.707$

$\cos 60{}^\circ =\dfrac{1}{2}=0.5$  

$\cos 90{}^\circ =0$ 

Therefore, we can conclude that the value of $\cos \theta $ decreases as $\theta $ increases. 

(iv) \[\sin \theta =\cos \theta \] for all values of \[\theta \].

Ans: The trigonometric ratio table is given as follows:

From the above table we can conclude that \[\sin \theta =\cos \theta \] is true only for $\theta =45{}^\circ $.

\[\sin \theta =\cos \theta \] is not true for all values of $\theta $.

(v) $\cot A$ is not defined for $A=0{}^\circ $.

Ans: We know that $\cot A=\dfrac{\cos A}{\sin A}$ .

If $A=0{}^\circ $, then $\cot 0{}^\circ =\dfrac{\cos 0{}^\circ }{\sin 0{}^\circ }$

From trigonometric ratio table we get

$\sin 0{}^\circ =0$ and $\cos 0{}^\circ =1$

$\cot 0{}^\circ =\dfrac{1}{0}$, which is undefined.

Exercise 8.3

(i) $\dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }$ 

Ans: The given expression is $\dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }$.

The given expression can be written as \[\dfrac{\sin \left( 90{}^\circ -72{}^\circ  \right)}{\cos 72{}^\circ }\].

Now, we can apply the identity $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $ , we get

$\dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=\dfrac{\sin \left( 90{}^\circ -72{}^\circ  \right)}{\cos 72{}^\circ }$

$\Rightarrow \dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=\dfrac{\cos 72{}^\circ }{\cos 72{}^\circ }$ 

$\therefore \dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=1$

(ii) $\dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }$ 

Ans: The given expression is $\dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }$.

The given expression can be written as $\dfrac{\tan \left( 90{}^\circ -64{}^\circ  \right)}{\cot 64{}^\circ }$.

Now, we can apply the identity $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $ , we get

$\dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=\dfrac{\tan \left( 90{}^\circ -64{}^\circ  \right)}{\cot 64{}^\circ }$

$\Rightarrow \dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=\dfrac{\cot 64{}^\circ }{\cot 64{}^\circ }$ 

$\therefore \dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=1$

(iii) $\cos 48{}^\circ -\sin 42{}^\circ $

Ans: The given expression is $\cos 48{}^\circ -\sin 42{}^\circ $.

The given expression can be written as $\cos \left( 90{}^\circ -42{}^\circ  \right)-\sin 42{}^\circ $.

Now, we can apply the identity $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $ , we get

$\cos 48{}^\circ -\sin 42{}^\circ =\cos \left( 90{}^\circ -42{}^\circ  \right)-\sin 42{}^\circ $

$\Rightarrow \cos 48{}^\circ -\sin 42{}^\circ =\sin 42{}^\circ -\sin 42{}^\circ $ 

$\therefore \cos 48{}^\circ -\sin 42{}^\circ =0$

(iv) $cosec 31{}^\circ -\sec 59{}^\circ $

Ans: The given expression is $cosec 31{}^\circ -\sec 59{}^\circ $.

The given expression can be written as $cosec\left( 90{}^\circ -59{}^\circ  \right)-\sec 59{}^\circ $.

Now, we can apply the identity $cosec\left( 90{}^\circ -\theta  \right)=\sec \theta $ , we get

$cosec 31{}^\circ -\sec 59{}^\circ =cosec\left( 90{}^\circ -59{}^\circ  \right)-\sec 59{}^\circ $

$\Rightarrow cosec31{}^\circ -\sec 59{}^\circ =\sec 59{}^\circ -\sec 59{}^\circ $ 

$\therefore cosec 31{}^\circ -\sec 59{}^\circ =0$

2. Show that 

(i) $\tan 48{}^\circ \tan 23{}^\circ \tan 42{}^\circ \tan 67{}^\circ =1$ 

Ans: The given expression is $\tan 48{}^\circ \tan 23{}^\circ \tan 42{}^\circ \tan 67{}^\circ =1$.

Let us consider LHS of the given expression, we get

$\tan 48{}^\circ \tan 23{}^\circ \tan 42{}^\circ \tan 67{}^\circ $

The above expression can be written as 

$\Rightarrow \tan \left( 90{}^\circ -42{}^\circ  \right)\tan \left( 90{}^\circ -67{}^\circ  \right)\tan 42{}^\circ \tan 67{}^\circ $

Now, we can apply the identity $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $, we get

$\Rightarrow \cot 42{}^\circ \cot 67{}^\circ \tan 42{}^\circ \tan 67{}^\circ $

Now, we know that $\cot A=\dfrac{1}{\tan A}$, we get

$\Rightarrow \dfrac{1}{\tan 42{}^\circ \tan 67{}^\circ }\times \tan 42{}^\circ \tan 67{}^\circ $

$\Rightarrow 1$ 

$\Rightarrow RHS$ 

$\therefore \tan 48{}^\circ \tan 23{}^\circ \tan 42{}^\circ \tan 67{}^\circ =1$ 

(ii) $\cos 38{}^\circ \cos 52{}^\circ -\sin 38{}^\circ \sin 52{}^\circ =0$ 

Ans: The given expression is $\cos 38{}^\circ \cos 52{}^\circ -\sin 38{}^\circ \sin 52{}^\circ =0$.

$\cos 38{}^\circ \cos 52{}^\circ -\sin 38{}^\circ \sin 52{}^\circ $

$\Rightarrow \cos \left( 90{}^\circ -52{}^\circ  \right)\cos \left( 90{}^\circ -38{}^\circ  \right)-\sin 38{}^\circ \sin 52{}^\circ $

Now, we can apply the identity $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $, we get

$\Rightarrow \sin 52{}^\circ \sin 38{}^\circ -\sin 38{}^\circ \sin 52{}^\circ $

$\Rightarrow 0$ 

$\therefore \cos 38{}^\circ \cos 52{}^\circ -\sin 38{}^\circ \sin 52{}^\circ =0$ 

3. Find the value of $A$, if $\tan 2A=\cot \left( A-18{}^\circ  \right)$, where $2A$ is an acute angle.

Ans: Given $\tan 2A=\cot \left( A-18{}^\circ  \right)$……….(1)

Now, we know that $\cot \left( 90{}^\circ -\theta  \right)=\tan \theta $.

Here, we can write $\tan 2A=\cot \left( 90{}^\circ -2A \right)$

Substitute the value in eq. (1), we get

$\Rightarrow \cot \left( 90{}^\circ -2A \right)=\cot \left( A-18{}^\circ  \right)$

Equating both angles, we get

$\Rightarrow \left( 90{}^\circ -2A \right)=\left( A-18{}^\circ  \right)$

$\Rightarrow 90{}^\circ +18{}^\circ =A+2A$ 

$\Rightarrow 108{}^\circ =3A$ 

$\Rightarrow 3A=108{}^\circ $

$\therefore A=36{}^\circ $ 

4. Prove that $A+B=90{}^\circ $, if $\tan A=\cot B$.

Ans: Given that $\tan A=\cot B$.

Now, substitute $\tan A=\cot \left( 90{}^\circ -A \right)$ in the given expression, we get

$\Rightarrow \cot \left( 90{}^\circ -A \right)=\cot B$

$\Rightarrow \left( 90{}^\circ -A \right)=B$

$\Rightarrow 90{}^\circ =B+A$

$\therefore A+B=90{}^\circ $

Hence proved

5. Find the value of $A$, if $\sec 4A=cosec\left( A-20{}^\circ  \right)$, where $4A$ is an acute angle.

Ans: Given $\sec 4A=cosec\left( A-20{}^\circ  \right)$……….(1)

Now, we know that $cosec\left( 90{}^\circ -\theta  \right)=\sec \theta $.

Here, we can write $\sec 4A=cosec\left( 90{}^\circ -4A \right)$

$\Rightarrow cosec\left( 90{}^\circ -4A \right)=cosec\left( A-20{}^\circ  \right)$

$\Rightarrow \left( 90{}^\circ -4A \right)=\left( A-20{}^\circ  \right)$

$\Rightarrow 90{}^\circ +20{}^\circ =A+4A$ 

$\Rightarrow 110{}^\circ =5A$ 

$\Rightarrow 5A=110{}^\circ $

$\therefore A=22{}^\circ $ 

6. If $A,B$ and $C$ are interior angles of a triangle $ABC$, then show that $\sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}$.

Ans: Given that $A,B$ and $C$ are interior angles of a triangle $ABC$.

We know that the sum of interior angles of a triangle is always $180{}^\circ $.

$\Rightarrow \angle A+\angle B+\angle C=180{}^\circ $ 

$\Rightarrow \angle B\angle C=180{}^\circ -\angle A$

Now, divide both sides of the equation by $2$, we get

$\Rightarrow \dfrac{\angle B+\angle C}{2}=\dfrac{180{}^\circ -\angle A}{2}$

$\Rightarrow \dfrac{\angle B+\angle C}{2}=90{}^\circ -\dfrac{\angle A}{2}$

Applying the sine function to the both sides of the equation, we get

$\Rightarrow \sin \left( \dfrac{\angle B+\angle C}{2} \right)=\sin \left( 90{}^\circ -\dfrac{\angle A}{2} \right)$

Now, we know that $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $.

$\therefore \sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}$.

7. Express $\sin 67{}^\circ +\cos 75{}^\circ $ in terms of trigonometric ratios of angles between $0{}^\circ $ and $45{}^\circ $.

Ans: Given expression $\sin 67{}^\circ +\cos 75{}^\circ $.

Now, we know that $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $.

The given expression can be written as 

$\sin 67{}^\circ +\cos 75{}^\circ =\cos \left( 90{}^\circ -23{}^\circ  \right)+\cos \left( 90{}^\circ -15{}^\circ  \right)$

$\therefore \sin 67{}^\circ +\cos 75{}^\circ =\cos 23{}^\circ +\cos 15{}^\circ $

Therefore, we get the expression in terms of trigonometric ratios of angles between $0{}^\circ $ and $45{}^\circ $.

Exercise 8.4

1. Express the trigonometric ratios $\sin A,\sec A$ and $\tan A$ in terms of $\cot A$.

Ans: For a right triangle we have an identity ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$.

Let us consider the above identity, we get

${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$

Now, reciprocating both sides we get

$\Rightarrow \dfrac{1}{{{\operatorname{cosec}}^{2}}A}=\dfrac{1}{1+{{\cot }^{2}}A}$

Now, we know that $\dfrac{1}{{{\operatorname{cosec}}^{2}}A}={{\sin }^{2}}A$, we get

$\Rightarrow {{\sin }^{2}}A=\dfrac{1}{1+{{\cot }^{2}}A}$

$\Rightarrow \sin A=\pm \dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$

Now, we know that sine value will be negative for angles greater than $180{}^\circ $, for a triangle sine value is always positive with respect to an angle. Then we will consider only positive values.

$\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$ 

We know that $\tan A=\dfrac{1}{\cot A}$ 

Also, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\sec }^{2}}A=1+{{\tan }^{2}}A$

$\Rightarrow {{\sec }^{2}}A=1+\dfrac{1}{{{\cot }^{2}}A}$

$\Rightarrow {{\sec }^{2}}A=\dfrac{{{\cot }^{2}}A+1}{{{\cot }^{2}}A}$

\[\Rightarrow \sec A=\dfrac{\sqrt{{{\cot }^{2}}A+1}}{\sqrt{{{\cot }^{2}}A}}\]

\[\therefore \sec A=\dfrac{\sqrt{{{\cot }^{2}}A+1}}{\cot A}\]

2. Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.

We know that $\cos A=\dfrac{1}{\sec A}$.

$\therefore \cos A=\dfrac{1}{\sec A}$

For a right triangle we have an identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.

${{\sin }^{2}}A+{{\cos }^{2}}A=1$

Now, we know that $\cos A=\dfrac{1}{\sec A}$, we get

$\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A$

$\Rightarrow {{\sin }^{2}}A=1-\dfrac{1}{{{\sec }^{2}}A}$

$\Rightarrow \sin A=\sqrt{1-{{\left( \dfrac{1}{\sec A} \right)}^{2}}}$

$\therefore \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}$

${{\tan }^{2}}A={{\sec }^{2}}A-1$

$\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}$ 

Now, we know that $\cot A=\dfrac{\cos A}{\sin A}$, we get

$\Rightarrow \cot A=\dfrac{\dfrac{1}{\sec A}}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}$

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$

We know that $cosecA=\dfrac{1}{\sin A}$, we get

 $\therefore cosecA=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}$

3. Evaluate the following:

(i) $\dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }$ 

Ans: The given expression is $\dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }$.

$\dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=\dfrac{{{\left[ \sin \left( 90{}^\circ -27{}^\circ  \right) \right]}^{2}}+{{\sin }^{2}}27{}^\circ }{{{\left[ \cos \left( 90{}^\circ -73{}^\circ  \right) \right]}^{2}}+{{\cos }^{2}}73{}^\circ }$

Now, we can apply the identity $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $ and $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $, we get

$\Rightarrow \dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=\dfrac{{{\cos }^{2}}27{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\sin }^{2}}73{}^\circ +{{\cos }^{2}}73{}^\circ }$

Now, by applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=\dfrac{1}{1}$ 

$\therefore \dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=1$  

(ii) $\sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ $ 

Ans: The given expression is $\sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ $ .

$\sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =\sin 25{}^\circ \cos \left( 90{}^\circ -25{}^\circ  \right)+\cos 25{}^\circ \sin \left( 90{}^\circ -25{}^\circ  \right)$

\[\Rightarrow \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =\sin 25{}^\circ \sin 25{}^\circ +\cos 25{}^\circ \cos 25{}^\circ \]

\[\Rightarrow \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ ={{\sin }^{2}}25{}^\circ +{{\cos }^{2}}25{}^\circ \]

\[\Rightarrow \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =1\] 

\[\therefore \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =1\]  

4. Choose the correct option and justify your choice:

(i) \[9{{\sec }^{2}}A-9{{\tan }^{2}}A=\] …….

(a) $1$ 

(c) $8$ 

Ans: The given expression is $9{{\sec }^{2}}A-9{{\tan }^{2}}A$.

$\Rightarrow 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)$

Now, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\sec }^{2}}A-{{\tan }^{2}}A=1$

$\Rightarrow 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9\left( 1 \right)$

$\therefore 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9$

Therefore, option (B) is the correct answer.

(ii) $\left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)$ 

Ans: The given expression is $\left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)$.

We know that the trigonometric functions have values as:

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$, $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$

Substituting these values in the given expression, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)\left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta } \right)$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta } \right)\left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta } \right)$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{{{\left( \sin \theta +\cos \theta  \right)}^{2}}-{{1}^{2}}}{\sin \theta \cos \theta }$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }$

$\therefore \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=2$

(iii) $\left( \sec A+\tan A \right)\left( 1-\sin A \right)=$ ………

(a) $\sec A$ 

(b) $\sin A$ 

(c) $cosecA$ 

(d) $\cos A$ 

Ans: Given expression is $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$.

We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$

$\left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)\left( 1-\sin A \right)$

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{1+\sin A}{\cos A} \right)\left( 1-\sin A \right)$

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{\left( 1+\sin A \right)\left( 1-\sin A \right)}{\cos A} \right)$

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{{{1}^{2}}-{{\sin }^{2}}A}{\cos A} \right)$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{{{\cos }^{2}}A}{\cos A} \right)$

$\therefore \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\cos A$

(iv) $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$ 

(a) ${{\sec }^{2}}A$ 

(b) $-1$ 

(c) ${{\cot }^{2}}A$ 

(d) ${{\tan }^{2}}A$ 

Ans: Given expression is $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$.

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

$\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{1+\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}}{1+\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{{{\cos }^{2}}A}}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\tan }^{2}}A$

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) ${{\left( cosec\theta -cot\theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$ 

Ans: Given expression is ${{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$.

Let us consider the LHS of the given expression, we get

$LHS={{\left( cosec\theta -\cot \theta  \right)}^{2}}$

Now, we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$ and $cosec\theta =\dfrac{1}{\sin \theta }$.

By substituting the values, we get

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}={{\left( \dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta } \right)}^{2}}$

 $\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}={{\left( \dfrac{1-\cos \theta }{\sin \theta } \right)}^{2}}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{{{\sin }^{2}}\theta }$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{1-{{\cos }^{2}}\theta }$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{\left( 1-\cos \theta  \right)\left( 1+\cos \theta  \right)}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{\left( 1-\cos \theta  \right)}{\left( 1+\cos \theta  \right)}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=RHS$

$\therefore {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

(ii) $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

Ans: Given expression is $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$.

$LHS=\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}$

Now, taking LCM, we get

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+\left( 1+\sin A \right)\left( 1+\sin A \right)}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A+2\sin A+1}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{1+2\sin A+1}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2+2\sin A}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2\left( 1+\sin A \right)}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2}{\cos A}$

We know that $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

\[\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=RHS\]

\[\therefore \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

(iii) $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta $ 

Ans: Given expression is $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta $.

$LHS=\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }$

Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }} \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{\sin \theta -\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{\dfrac{\cos \theta -\sin \theta }{\cos \theta }} \right)\]

 \[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta \left( \sin \theta -\cos \theta  \right)}+\dfrac{{{\cos }^{2}}\theta }{\sin \theta \left( \sin \theta -\cos \theta  \right)} \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta }+\dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left( \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta \cos \theta } \right)\]

Now, by applying the identity \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\], we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left[ \dfrac{\left( \sin \theta -\cos \theta  \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta } \right]\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left[ \dfrac{\left( \sin \theta -\cos \theta  \right)\left( 1+\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta } \right]\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{\left( 1+\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\sin \theta \cos \theta }+\dfrac{\sin \theta \cos \theta }{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\sin \theta \cos \theta }+1\]

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\sec \theta cosec\theta +1\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta \]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=RHS\]

\[\therefore \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta \]

(iv) $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$ 

Ans: Given expression is $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$.

$LHS=\dfrac{1+\sec A}{\sec A}$

Now, we know that $\sec \theta =\dfrac{1}{\cos \theta }$.

By substituting the value, we get

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{\dfrac{\cos A+1}{\cos A}}{\dfrac{1}{\cos A}}\]

 \[\Rightarrow \dfrac{1+\sec A}{\sec A}=\cos A+1\]

Multiply and divide by $\left( 1-\cos A \right)$, we get

 \[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{\left( 1+\cos A \right)\left( 1-\cos A \right)}{\left( 1-\cos A \right)}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{1-{{\cos }^{2}}A}{\left( 1-\cos A \right)}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{\left( 1-\cos A \right)}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=RHS\]

\[\therefore \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

(v) $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$ 

Ans: Given expression is $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$.

Now, let us consider the LHS of the given expression, we get

$LHS=\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}$

Dividing numerator and denominator by $\sin A$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\sin A}{\sin A}+\dfrac{1}{\sin A}}{\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\sin A}-\dfrac{1}{\sin A}}$

Now, we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$ and $cosec\theta =\dfrac{1}{\sin \theta }$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-1+\operatorname{cosec}A}{\cot A+1-\operatorname{cosec}A}$

Now, by applying the identity ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$, substitute $1={{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-\left( {{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A \right)+\operatorname{cosec}A}{\cot A+{{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A-\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-{{\cot }^{2}}A+{{\operatorname{cosec}}^{2}}A+\operatorname{cosec}A}{\cot A+{{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A-\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{{{\left( \cot A-1+\operatorname{cosec}A \right)}^{2}}}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2{{\operatorname{cosec}}^{2}}A+2\cot A\operatorname{cosec}A-2\cot A-2\operatorname{cosec}A}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2\operatorname{cosec}A\left( \cot A-\operatorname{cosec}A \right)-2\left( \cot A-\operatorname{cosec}A \right)}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\left( 2\operatorname{cosec}A-2 \right)\left( \cot A-\operatorname{cosec}A \right)}{1-1+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\left( 2\operatorname{cosec}A-2 \right)\left( \cot A-\operatorname{cosec}A \right)}{2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec}A+\cot A$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=RHS$

$\therefore \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec}A+\cot A$ 

(vi) $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$ 

Ans: Given expression is $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$.

$LHS=\sqrt{\dfrac{1+\sin A}{1-\sin A}}$

Now, multiply and divide the expression by $\sqrt{1+\sin A}$, we get

$\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sqrt{\dfrac{\left( 1+\sin A \right)\left( 1+\sin A \right)}{\left( 1-\sin A \right)\left( 1+\sin A \right)}}$

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sqrt{\dfrac{{{\left( 1+\sin A \right)}^{2}}}{1-{{\sin }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1+\sin A}{\sqrt{{{\cos }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1+\sin A}{\cos A}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=RHS\]

$\therefore \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$ 

(vii) $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $ 

Ans: Given expression is $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $.

$LHS=\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }$

Taking common terms out, we get

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 2\left( 1-2{{\sin }^{2}}\theta  \right)-1 \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 2-2{{\sin }^{2}}\theta -1 \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 1-2{{\sin }^{2}}\theta  \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta }{\cos \theta }$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=RHS$

$\therefore \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $

(viii) ${{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+secA \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$ 

Ans: Given expression is ${{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$.

$LHS={{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}$

Now, by applying the identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], we get

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}={{\sin }^{2}}A+cosec{{A}^{2}}+2\sin AcosecA+{{\cos }^{2}}A+{{\sec }^{2}}A+2\cos A\sec A\]\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}={{\sin }^{2}}A+{{\cos }^{2}}A+cosec{{A}^{2}}+{{\sec }^{2}}A+2\sin AcosecA+2\cos A\sec A\]We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=1+cose{{c}^{2}}\theta +{{\sec }^{2}}\theta +2\sin A\dfrac{1}{\sin A}+2\cos A\dfrac{1}{\cos A}\] 

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=1+\left( 1+{{\cot }^{2}}A+1+{{\tan }^{2}}A \right)+2+2\]

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=RHS\]

\[\therefore {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

(ix) $\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$

Ans: Given expression is $\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$.

$LHS=\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\left( \dfrac{{{\sin }^{2}}A}{\cos A} \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\sin A\cos A$

Now, consider the RHS of the given expression, we get

$RHS=\dfrac{1}{\tan A+\cot A}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{1}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A}}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{\sin A\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\sin A\cos A$

Here, we get LHS=RHS

$\therefore \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$ 

(x) $\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$

Ans: Given expression is $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$.

$LHS=\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$

By applying the identities ${{\sec }^{2}}A=1+{{\tan }^{2}}A$ and ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$, we get

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{se{{c}^{2}}A}{{{\operatorname{cosec}}^{2}}A}$

$RHS={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$

Now, we know that $\cot \theta =\dfrac{1}{\tan \theta }$, we get

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{1-\dfrac{1}{\tan A}} \right)}^{2}}$

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{\dfrac{\tan A-1}{\tan A}} \right)}^{2}}$

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( -\tan A \right)}^{2}}$

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$

$\therefore \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$ 

NCERT Solutions for Class 10 Maths Chapter 8 - Free PDF Download

We at Vedantu will help students overcome all difficulties and get the best possible marks in their examinations. We provide students with a PDF of NCERT questions that are solved. All these exercise questions will allow them to understand the kinds of sums frequently asked in the examination.

You can find the Solutions of All the Class 10 Maths Chapters below.

NCERT Solutions for Class 10 Maths

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Pair of Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometry

Chapter 8 - Introduction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Constructions

Chapter 12 - Areas Related to Circles

Chapter 13 - Surface Areas and Volumes

Chapter 14 - Statistics

Chapter 15 - Probability

About the Chapter - Introduction to Trigonometry

Trigonometry is all about triangles. It is all about right-angled triangles, triangles with one angle equal to 90 degrees, to be more precise. It's a method that helps us find a triangle's missing angles and missing sides. The ‘trigono’ word means triangle and the ‘metry’ word means to measure. 

Chapter 8 Introduction to Class 10 Trigonometry NCERT Syllabus is divided into five parts and four exercises. The last part of the exercise consists of problems that can be pictured using the right angle triangle. The second section consists of an introduction to trigonometric ratios with examples. It is accompanied by an exercise at the end of the section and the derivation of sine, cosine, and other trigonometric functions.  

The third part consists of trigonometric ratios about measurement. The fourth section of Chapter 8 Class 10 Maths consists of a few solved examples, trigonometric ratio criteria for complementary angles, and an exercise. In the introduction to Trigonometry Class 10, the fifth section consists of the subject relating to trigonometric identities, with a few examples, and ends with an exercise.

Key Concepts at a Glance

The following are the key concepts that are covered in the chapter.

Opposite & Adjacent Sides in a Right Angled Triangle

The sides of a right-angle triangle are referred to as opposite, adjacent, and hypotenuse.

The hypotenuse is the side that is opposite to the right angle. It's also the triangle's longest side.

Adjacent and opposite sides are in respect to the angle specified and might vary accordingly. The opposite side is the side that is opposite the specified angle, while the adjacent side is the side that is right by the specified angle.

Basic Trigonometric Ratios

In ΔABC, right-angled at ∠B, the trigonometric ratios of the ∠A are as follows:

sin A=opposite side/hypotenuse=BC/AC

cos A=adjacent side/hypotenuse=AB/AC

tan A=opposite side/adjacent side=BC/AB

cosec A=hypotenuse/opposite side=AC/BC

sec A=hypotenuse/adjacent side=AC/AB

cot A=adjacent side/opposite side=AB/BC

Standard Values of Trigonometric Ratios

Complementary trigonometric ratios:.

sin (90 o − θ) = cos θ

cos (90 o − θ) = sin θ

tan (90 o − θ) = cot θ

cot (90 o − θ) = tan θ

cosec (90 o − θ) = sec θ

sec (90 o − θ) = cosec θ

Significance of NCERT Solutions for Class 10 Maths Chapters 8- Introducing to Trigonometry

The NCERT Solutions for Class 10 Maths Chapter 8 - "Introduction to Trigonometry" hold immense significance in the academic journey of students. This chapter serves as a gateway to understanding the fundamental principles of trigonometry, a branch of mathematics with wide-ranging applications. The solutions provide a structured and comprehensive approach to mastering concepts such as trigonometric ratios, sine, cosine, tangent, and their interplay within triangles and angles. By offering step-by-step explanations and detailed solutions to exercises, these resources aid students in grasping complex ideas with clarity. Moreover, the practical applications of trigonometry, showcased in real-world scenarios like measuring heights and distances, highlight the subject's relevance beyond the classroom.

The significance of these solutions extends to exam preparation, as they align closely with the NCERT curriculum, a key reference for board exams. Students can practice a variety of questions that mimic exam formats, enhancing their problem-solving skills and boosting their confidence. Furthermore, the visual aids and diagrams included in the solutions facilitate visual learning, helping students comprehend abstract concepts more easily. As a foundational topic, mastering trigonometry is essential for students progressing to advanced mathematical concepts in higher classes. Overall, the NCERT Solutions for Chapter 8 provide an indispensable resource that not only nurtures mathematical competence but also nurtures critical thinking and analytical skills, fostering a strong educational foundation for students.

Benefits of NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths provides many benefits. Some of the benefits and features of these NCERT Strategies are:

To save students from the needless burden on their minds, complicated solutions are broken down into simple sections.

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To facilitate the understanding of the concept, topics and answers are incorporated with relevant illustrations. 

All the solutions are provided in adherence to the guidelines set by CBSE. 

In the chapter Introduction to Trigonometry, these NCERT Solutions are precisely written so that students do not face any difficulties while solving any questions. 

Our experts have made sure that it is straightforward to understand these NCERT solutions. 

In the form of solutions, the PDF encompasses the entire syllabus and theory.

The solutions are provided with the new syllabus and exam requirements in mind.

All study materials, and NCERT solutions are available on the website and app of Vedantu. All the NCERT solutions are available for free download on Vedantu in PDF format. Along with this, students can also view additional study materials provided by Vedantu, for Class 10 Maths 

CBSE Class 10 Maths Chapter Wise Formulas

CBSE Class 10 Maths Revision Notes

CBSE Class 10 Maths Syllabus

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Conclusion  

NCERT Solutions for Class 10 Maths Chapter 8 - "Introduction to Trigonometry" provides a comprehensive understanding of fundamental trigonometric concepts. This chapter lays the foundation for the study of angles, triangles, and their relationships within trigonometry. By addressing topics such as ratios, sine, cosine, tangent, and applications in real-world scenarios, these solutions equip students with the tools to comprehend and solve trigonometric problems. The clear explanations and step-by-step solutions enhance students' mathematical proficiency and problem-solving skills. This chapter not only aids exam preparation but also instils a solid grasp of trigonometry, facilitating a smoother transition to more advanced mathematical concepts.

FAQs on NCERT Solutions for Class 10 Maths Chapter 8- Introduction To Trigonometry

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3. What do you learn about trigonometry in Class 10 Maths Chapter 8?

Chapter 8 of Class 10 Maths is basically about trigonometry which is a very important topic for Class 10. It covers the introduction to the ratios and identities, trigonometric ratios of some specific angles, ratios of some complementary angles, and trigonometric identities to solve equations. Visit Vedantu’s website(vedantu.com) to take a look at the solutions for all the exercises. You can refer to Vedantu’s NCERT Solutions for Chapter 8 of Class 10 Maths Trigonometry to understand the concepts better and score high marks in exams. These solutions are available at free of cost on Vedantu website and mobile app.

4. What is the objective of Chapter 8 in Class 10 Maths Trigonometry?

The main objective of Chapter 8 of Class 10 Maths is to introduce a very important concept to students. Trigonometry is a key concept which will play a role in higher studies and also be a part of Physics numerical problems. Students should make sure that the basics of this chapter are understood well. To get an in-depth understanding of the concepts, refer to NCERT solutions, revision notes and important questions offered by Vedantu. These are available free of cost.

5. What does Exercise 8.4 of Chapter 8 Trigonometry of Class 10 Maths  deal with?

Exercise 8.4 of Chapter 8 Trigonometry of Class 10 Maths deals with trigonometric ratios of complementary angles. Complementary angles are those angles whose sum adds up to a total of 90 degrees. There are some standard formulae which are important to be memorised. Students are advised to solve every single question from this exercise and refer to Vedantu’s solutions for further answers.

NCERT Solutions for Class 10

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• Chapter 8: Introduction To Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Ncert solutions for class 10 maths chapter 8 – download free pdf.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry is helpful for the students as it aids in understanding the concepts as well as in scoring well in CBSE Class 10 board examination. The NCERT Solutions are designed and reviewed by subject experts and cover all the questions from the textbook. These NCERT Solutions are framed as per the latest update on the CBSE Syllabus for 2023-24 and its guidelines, in accordance with the exam pattern.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 8 Introduction to Trigonometry

Download most important questions for class 10 maths chapter – 8 introduction to trigonometry.

The  NCERT Solutions for Class 10 Maths provide a strong foundation for every concept across all chapters. Students can clarify their doubts and understand the fundamentals of this chapter. Also, students can solve the difficult problems in each exercise with the help of these NCERT Solutions for Class 10  Maths Chapter 8.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability

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Exercise 8.1 Page: 181

1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A

(ii) sin C, cos C

In a given triangle ABC, right angled at B = ∠B = 90°

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem, we get

AC 2 =AB 2 +BC 2

AC 2  = (24) 2 +7 2

AC 2  = (576+49)

AC 2  = 625cm 2

AC = √625 = 25

Therefore, AC = 25 cm

(i) To find Sin (A), Cos (A)

We know that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,

Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25

(ii) To find Sin (C), Cos (C)

Sin (C) = AB/AC = 24/25

Cos (C) = BC/AC = 7/25

2. In Fig. 8.13, find tan P – cot R

In the given triangle PQR, the given triangle is right angled at Q and the given measures are:

Since the given triangle is right angled triangle, to find the side QR, apply the Pythagorean theorem

According to Pythagorean theorem,

PR 2 = QR 2 + PQ 2

Substitute the values of PR and PQ

13 2 = QR 2 +12 2

169 = QR 2 +144

Therefore, QR 2 = 169−144

QR = √25 = 5

Therefore, the side QR = 5 cm

To find tan P – cot R:

According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes

tan (P) = Opposite side /Adjacent side = QR/PQ = 5/12

Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,

Cot (R) = Adjacent side/Opposite side = QR/PQ = 5/12

tan (P) – cot (R) = 5/12 – 5/12 = 0

Therefore, tan(P) – cot(R) = 0

3. If sin A = 3/4, calculate cos A and tan A.

Let us assume a right angled triangle ABC, right angled at B

Given: Sin A = 3/4

We know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.

Therefore, Sin A = Opposite side /Hypotenuse= 3/4

Let BC be 3k and AC will be 4k

where k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC 2 =AB 2 + BC 2

Substitute the value of AC and BC

(4k) 2 =AB 2 + (3k) 2

16k 2 −9k 2 =AB 2

Therefore, AB = √7k

Now, we have to find the value of cos A and tan A

We know that,

Cos (A) = Adjacent side/Hypotenuse

Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get

AB/AC = √7k/4k = √7/4

Therefore, cos (A) = √7/4

tan(A) = Opposite side/Adjacent side

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

BC/AB = 3k/√7k = 3/√7

Therefore, tan A = 3/√7

4. Given 15 cot A = 8, find sin A and sec A.

Given: 15 cot A = 8

So, Cot A = 8/15

We know that, cot function is the equal to the ratio of length of the adjacent side to the opposite side.

Therefore, cot A = Adjacent side/Opposite side = AB/BC = 8/15

Let AB be 8k and BC will be 15k

Where, k is a positive real number.

Substitute the value of AB and BC

AC 2 = (8k) 2 + (15k) 2

AC 2 = 64k 2 + 225k 2

AC 2 = 289k 2

Therefore, AC = 17k

Now, we have to find the value of sin A and sec A

Sin (A) = Opposite side /Hypotenuse

Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get

Sin A = BC/AC = 15k/17k = 15/17

Therefore, sin A = 15/17

Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side.

Sec (A) = Hypotenuse/Adjacent side

AC/AB = 17k/8k = 17/8

Therefore sec (A) = 17/8

5. Given sec θ = 13/12 Calculate all other trigonometric ratios

We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side

sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB

Let AC be 13k and AB will be 12k

Substitute the value of AB and AC

(13k) 2 = (12k) 2 + BC 2

169k 2 = 144k 2 + BC 2

BC 2 = 169k 2 – 144k 2

BC 2 = 25k 2

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13

Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13

tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12

Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5

cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Let us assume the triangle ABC in which CD⊥AB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value

AD/BD = AC/BC = k

Now consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

CD 2 = BC 2 – BD 2 … (3)

CD 2 =AC 2 −AD 2  ….(4)

From the equations (3) and (4) we get,

AC 2 −AD 2 = BC 2 −BD 2

Now substitute the equations (1) and (2) in (3) and (4)

K 2 (BC 2 −BD 2 )=(BC 2 −BD 2 ) k 2 =1

Putting this value in equation, we obtain

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)

7. If cot θ = 7/8, evaluate :

(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)

(ii) cot 2 θ

Let us assume a △ABC in which ∠B = 90° and ∠C = θ

cot θ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number

According to Pythagoras theorem in △ABC we get.

AC 2 = AB 2 +BC 2

AC 2 = (8k) 2 +(7k) 2

AC 2 = 64k 2 +49k 2

AC 2 = 113k 2

AC = √113 k

According to the sine and cos function ratios, it is written as

sin θ = AB/AC = Opposite Side/Hypotenuse = 8k/√113 k = 8/√113 and

cos θ = Adjacent Side/Hypotenuse = BC/AC = 7k/√113 k = 7/√113

Now apply the values of sin function and cos function:

8. If 3 cot A = 4, check whether (1-tan 2 A)/(1+tan 2 A) = cos 2 A – sin 2 A or not.

Let △ABC in which ∠B=90°

We know that, cot function is the reciprocal of tan function and it is written as

cot(A) = AB/BC = 4/3

Let AB = 4k an BC =3k, where k is a positive real number.

According to the Pythagorean theorem,

AC 2 =(4k) 2 +(3k) 2

AC 2 =16k 2 +9k 2

AC 2 =25k 2

Now, apply the values corresponding to the ratios

tan(A) = BC/AB = 3/4

sin (A) = BC/AC = 3/5

cos (A) = AB/AC = 4/5

Now compare the left hand side(LHS) with right hand side(RHS)

Since, both the LHS and RHS = 7/25

R.H.S. =L.H.S.

Hence, (1-tan 2 A)/(1+tan 2 A) = cos 2 A – sin 2 A   is proved

9. In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Let ΔABC in which ∠B=90°

tan A = BC/AB = 1/√3

Let BC = 1k and AB = √3 k,

Where k is the positive real number of the problem

By Pythagoras theorem in ΔABC we get:

AC 2 =(√3 k) 2 +(k) 2

AC 2 =3k 2 +k 2

Now find the values of cos A, Sin A

Sin A = BC/AC = 1/2

Cos A = AB/AC = √3/2

Then find the values of cos C and sin C

Sin C = AB/AC = √ 3/2

Cos C = BC/AC = 1/2

Now, substitute the values in the given problem

(i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1

(ii) cos A cos C – sin A sin C = ( √ 3/2 )(1/2) – (1/2) ( √ 3/2 ) = 0

10. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

In a given triangle PQR, right angled at Q, the following measures are

PR + QR = 25 cm

Now let us assume, QR = x

According to the Pythagorean Theorem,

PR 2 = PQ 2 + QR 2

Substitute the value of PR as x

(25- x) 2 = 5 2 + x 2

25 2 + x 2 – 50x = 25 + x 2

625 + x 2 -50x -25 – x 2 = 0

-50x = -600

x= -600/-50

x = 12 = QR

Now, find the value of PR

PR = 25- QR

Substitute the value of QR

Now, substitute the value to the given problem

(1) sin p = Opposite Side/Hypotenuse = QR/PR = 12/13

(2) Cos p = Adjacent Side/Hypotenuse = PQ/PR = 5/13

(3) tan p =Opposite Side/Adjacent side = QR/PQ = 12/5

11. State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii)cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 4/3 for some angle θ.

Answer: False

Proof: In ΔMNC in which ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 which is greater than 1.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.

MC 2 =MN 2 +NC 2

5 2 =3 2 +4 2

25   =   25

(ii) sec A = 12/5 for some value of angle A

Answer: True

Justification: Let a ΔMNC in which ∠N = 90º,

MC=12k and MB=5k, where k is a positive real number.

By Pythagoras theorem we get,

(12k) 2 =(5k) 2 +NC 2

NC 2 +25k 2 =144k 2

NC 2 =119k 2

Such a triangle is possible as it will follow the Pythagoras theorem.

(iii) cos A is the abbreviation used for the cosecant of angle A.

Justification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.

Justification: cot M is not the product of cot and M. It is the cotangent of ∠M.

Answer : False

Justification: sin θ = Opposite/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.

Exercise 8.2 Page: 187

1. Evaluate the following:

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan 2 45° + cos 2 30° – sin 2 60

First, find the values of the given trigonometric ratios

sin 30° = 1/2

cos 30° = √3/2

sin 60° = 3/2

cos 60°= 1/2

sin 60° cos 30° + sin 30° cos 60° = √3/2 ×√3/2 + (1/2) ×(1/2 ) = 3/4+1/4 = 4/4 =1

We know that, the values of the trigonometric ratios are:

sin 60° = √3/2

tan 45° = 1

Substitute the values in the given problem

2 tan 2 45° + cos 2 30° – sin 2 60 = 2(1) 2 + (√3/2) 2 -(√3/2) 2

2 tan 2 45° + cos 2 30° – sin 2 60 = 2 + 0

2 tan 2 45° + cos 2 30° – sin 2 60 = 2

(iii) cos 45°/(sec 30°+cosec 30°)

cos 45° = 1/√2

sec 30° = 2/√3

cosec 30° = 2

Substitute the values, we get

Now, multiply both the numerator and denominator by √2 , we get

Therefore, cos 45°/(sec 30°+cosec 30°) = (3√2 – √6)/8

cosec 60° = 2/√3

cos 60° = 1/2

cot 45° = 1

Substitute the values in the given problem, we get

Now, substitute the values in the given problem, we get

(5cos 2 60° + 4sec 2 30° – tan 2 45°)/(sin 2 30° + cos 2 30°)

= 5(1/2) 2 +4(2/√3) 2 -1 2 /(1/2) 2 +(√3/2) 2

  = (5/4+16/3-1)/(1/4+3/4)

= (15+64-12)/12/(4/4)

2. Choose the correct option and justify your choice : (i) 2tan 30°/1+tan 2 30° = (A) sin 60°            (B) cos 60°          (C) tan 60°            (D) sin 30° (ii) 1-tan 2 45°/1+tan 2 45° = (A) tan 90°            (B) 1                    (C) sin 45°            (D) 0 (iii)  sin 2A = 2 sin A is true when A = (A) 0°                   (B) 30°                  (C) 45°                 (D) 60°

(iv) 2tan30°/1-tan 2 30° = (A) cos 60°          (B) sin 60°             (C) tan 60°           (D) sin 30°

(i) (A) is correct.

Substitute the of tan 30° in the given equation

tan 30° = 1/√3

2tan 30°/1+tan 2 30° = 2(1/√3)/1+(1/√3) 2

= (2/√3)/(1+1/3) = (2/√3)/(4/3)

= 6/4√3 = √3/2 = sin 60°

The obtained solution is equivalent to the trigonometric ratio sin 60°

(ii) (D) is correct.

Substitute the of tan 45° in the given equation

1-tan 2 45°/1+tan 2 45° = (1-1 2 )/(1+1 2 )

The solution of the above equation is 0.

(iii) (A) is correct.

To find the value of A, substitute the degree given in the options one by one

sin 2A = 2 sin A is true when A = 0°

As sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 2 × 0 = 0

Apply the sin 2A formula, to find the degree value

sin 2A = 2sin A cos A

⇒2sin A cos A = 2 sin A

⇒ 2cos A = 2 ⇒ cos A = 1

Now, we have to check, to get the solution as 1, which degree value has to be applied.

When 0 degree is applied to cos value, i.e., cos 0 =1

Therefore, ⇒ A = 0°

(iv) (C) is correct.

2tan30°/1-tan 2 30° =  2(1/√3)/1-(1/√3) 2

= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°

The value of the given equation is equivalent to tan 60°.

3. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.

tan (A + B) = √3

Since √3 = tan 60°

Now substitute the degree value

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) = 1/√3

Since 1/√3 = tan 30°

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

Therefore A = 45° and B = 15°

4. State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

Justification:

Let us take A = 30° and B = 60°, then

Substitute the values in the sin (A + B) formula, we get

sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,

sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = 1+√3/2

Since the values obtained are not equal, the solution is false.

According to the values obtained as per the unit circle, the values of sin are:

sin 45° = 1/√2

sin 60° = √3/2

sin 90° = 1

Thus the value of sin θ increases as θ increases. Hence, the statement is true

(iii) False.

According to the values obtained as per the unit circle, the values of cos are:

cos 60° = 1/2

cos 90° = 0

Thus, the value of cos θ decreases as θ increases. So, the statement given above is false.

sin θ = cos θ, when a right triangle has 2 angles of (π/4). Therefore, the above statement is false.

Since cot function is the reciprocal of the tan function, it is also written as:

cot A = cos A/sin A

Now substitute A = 0°

cot 0° = cos 0°/sin 0° = 1/0 = undefined.

Hence, it is true

Exercise 8.3 Page: 189

1. Evaluate :

(i) sin 18°/cos 72°        

(ii) tan 26°/cot 64°      

(iii)  cos 48° – sin 42°      

(iv)  cosec 31° – sec 59°

(i) sin 18°/cos 72°

To simplify this, convert the sin function into cos function

We know that, 18° is written as 90° – 18°, which is equal to the cos 72°.

= sin (90° – 18°) /cos 72°

Substitute the value, to simplify this equation

= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°

To simplify this, convert the tan function into cot function

We know that, 26° is written as 90° – 26°, which is equal to the cot 64°.

= tan (90° – 26°)/cot 64°

= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°

To simplify this, convert the cos function into sin function

We know that, 48° is written as 90° – 42°, which is equal to the sin 42°.

= cos (90° – 42°) – sin 42°

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

To simplify this, convert the cosec function into sec function

We know that, 31° is written as 90° – 59°, which is equal to the sec 59°

= cosec (90° – 59°) – sec 59°

= sec 59° – sec 59° = 0

2.  Show that:

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

(i) tan 48° tan 23° tan 42° tan 67°

Simplify the given problem by converting some of the tan functions to the cot functions

We know that, tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute the values

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°

Simplify the given problem by converting some of the cos functions to the sin functions

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°

= sin 52° sin 38° – sin 38° sin 52° = 0

3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A .

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

Substitute the above equation in the given problem

⇒ cot (90° – 2A) = cot (A -18°)

Now, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

Therefore, the value of A = 36°

4.  If tan A = cot B, prove that A + B = 90°.

tan A = cot B

We know that cot B = tan (90° – B)

To prove A + B = 90°, substitute the above equation in the given problem

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

Hence Proved.

5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

sec 4A = cosec (A – 20°)

We know that sec 4A = cosec (90° – 4A)

To find the value of A, substitute the above equation in the given problem

cosec (90° – 4A) = cosec (A – 20°)

Now, equate the angles

90° – 4A= A- 20°

A = 110°/ 5 = 22°

Therefore, the value of A = 22°

6. If A, B and C are interior angles of a triangle ABC, then show that

    sin (B+C/2) = cos A/2

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

Now, multiply both sides by sin functions, we get

⇒ sin (B+C)/2 = sin (90°-A/2)

Since sin (90°-A/2) = cos A/2, the above equation is equal to

sin (B+C)/2 = cos A/2

Hence proved.

7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

sin 67° + cos 75°

In term of sin as cos function and cos as sin function, it can be written as follows

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

So, sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)

Now, simplify the above equation

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is also expressed as cos 23° + sin 15°

Exercise 8.4 Page: 193

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas

cosec 2 A   – cot 2 A = 1

cosec 2 A = 1 + cot 2 A

Since cosec function is the inverse of sin function, it is written as

1/sin 2 A = 1 + cot 2 A

Now, rearrange the terms, it becomes

sin 2 A = 1/(1+cot 2 A)

Now, take square roots on both sides, we get

sin A = ±1/(√(1+cot 2 A)

The above equation defines the sin function in terms of cot function

Now, to express sec function in terms of cot function, use this formula

sin 2 A = 1/ (1+cot 2 A)

Now, represent the sin function as cos function

1 – cos 2 A = 1/ (1+cot 2 A)

Rearrange the terms,

cos 2 A = 1 – 1/(1+cot 2 A)

⇒cos 2 A = (1-1+cot 2 A)/(1+cot 2 A)

Since sec function is the inverse of cos function,

⇒ 1/sec 2 A = cot 2 A/(1+cot 2 A)

Take the reciprocal and square roots on both sides, we get

⇒ sec A = ±√ (1+cot 2 A)/cotA

Now, to express tan function in terms of cot function

tan A = sin A/cos A and cot A = cos A/sin A

Since cot function is the inverse of tan function, it is rewritten as

tan A = 1/cot A

2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution:

Cos A function in terms of sec A:

sec A = 1/cos A

⇒ cos A = 1/sec A

sec A function in terms of sec A:

cos 2 A + sin 2 A = 1

Rearrange the terms

sin 2 A = 1 – cos 2 A

sin 2 A = 1 – (1/sec 2 A)

sin 2 A = (sec 2 A-1)/sec 2 A

sin A = ± √(sec 2 A-1)/sec A

cosec A function in terms of sec A:

sin A = 1/cosec A

⇒cosec A = 1/sin A

cosec A = ± sec A/√(sec 2 A-1)

Now, tan A function in terms of sec A:

sec 2 A – tan 2 A = 1

⇒ tan 2 A = sec 2 A – 1

tan A = √(sec 2 A – 1)

cot A function in terms of sec A:

⇒ cot A = 1/tan A

cot A = ±1/√(sec 2 A – 1)

3. Evaluate:

(i) (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°) (ii)  sin 25° cos 65° + cos 25° sin 65°

(i) (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°)

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= [sin 2 (90°-27°) + sin 2 27°] / [cos 2 (90°-73°) + cos 2 73°)]

= (cos 2 27°   + sin 2 27°)/(sin 2 27° + cos 2 73°)

= 1/1 =1                       (since sin 2 A + cos 2 A = 1)

Therefore, (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°) = 1

(ii) sin 25° cos 65° + cos 25° sin 65°

= sin(90°-25°) cos 65° + cos (90°-65°) sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= cos 2 65°   + sin 2 65° = 1 (since sin 2 A + cos 2 A = 1)

Therefore, sin 25° cos 65° + cos 25° sin 65° = 1

4. Choose the correct option. Justify your choice. (i) 9 sec 2 A – 9 tan 2 A = (A) 1                 (B) 9              (C) 8                (D) 0 (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (A) 0                 (B) 1              (C) 2                (D) – 1 (iii) (sec A + tan A) (1 – sin A) = (A) sec A           (B) sin A        (C) cosec A      (D) cos A

(iv) 1+tan 2 A/1+cot 2 A = 

      (A) sec 2 A                 (B) -1              (C) cot 2 A                (D) tan 2 A

(i) (B) is correct.

Take 9 outside, and it becomes

9 sec 2 A – 9 tan 2 A

= 9 (sec 2 A – tan 2 A)

= 9×1 = 9             (∵ sec2 A – tan2 A = 1)

Therefore, 9 sec 2 A – 9 tan 2 A = 9

(ii) (C) is correct

(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

We know that, tan θ = sin θ/cos θ

sec θ = 1/ cos θ

cot θ = cos θ/sin θ

cosec θ = 1/sin θ

Now, substitute the above values in the given problem, we get

= (1 + sin θ/cos θ + 1/ cos θ) (1 + cos θ/sin θ – 1/sin θ)

Simplify the above equation,

= (cos θ +sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ

= (cos θ+sin θ) 2 -1 2 /(cos θ sin θ)

= (cos 2 θ + sin 2 θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos 2 θ + sin 2 θ = 1)

= (2cos θ sin θ)/(cos θ sin θ) = 2

Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2

(iii) (D) is correct.

Sec A= 1/cos A

Tan A = sin A / cos A

(secA + tanA) (1 – sinA)

= (1/cos A + sin A/cos A) (1 – sinA)

= (1+sin A/cos A) (1 – sinA)

= (1 – sin 2 A)/cos A

= cos 2 A/cos A = cos A

Therefore, (secA + tanA) (1 – sinA) = cos A

(iv) (D) is correct.

tan 2 A =1/cot 2 A

Now, substitute this in the given problem, we get

1+tan 2 A/1+cot 2 A

= (1+1/cot 2 A)/1+cot 2 A

= (cot 2 A+1/cot 2 A)×(1/1+cot 2 A)

= 1/cot 2 A = tan 2 A

So, 1+tan 2 A/1+cot 2 A = tan 2 A

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec θ – cot θ) 2  = (1-cos θ)/(1+cos θ)

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

     [Hint : Write the expression in terms of sin θ and cos θ]

(iv) (1 + sec A)/sec A = sin 2 A/(1-cos A)  

     [Hint : Simplify LHS and RHS separately]

(v) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec 2 A = 1+cot 2 A.

(vii) (sin θ – 2sin 3 θ)/(2cos 3 θ-cos θ) = tan θ (viii) (sin A + cosec A) 2  + (cos A + sec A) 2  = 7+tan 2 A+cot 2 A (ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA) [Hint : Simplify LHS and RHS separately] (x) (1+tan 2 A/1+cot 2 A) = (1-tan A/1-cot A) 2  =   tan 2 A

To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)

L.H.S. = (cosec θ – cot θ) 2

The above equation is in the form of (a-b) 2 , and expand it

Since (a-b) 2 = a 2 + b 2 – 2ab

Here a = cosec θ and b = cot θ

= (cosec 2 θ + cot 2 θ – 2cosec θ cot θ)

Now, apply the corresponding inverse functions and equivalent ratios to simplify

= (1/sin 2 θ + cos 2 θ/sin 2 θ – 2cos θ/sin 2 θ)

= (1 + cos 2 θ – 2cos θ)/(1 – cos 2 θ)

= (1-cos θ) 2 /(1 – cosθ)(1+cos θ)

= (1-cos θ)/(1+cos θ) = R.H.S.

Therefore, (cosec θ – cot θ) 2  = (1-cos θ)/(1+cos θ)

(ii)  (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

Now, take the L.H.S of the given equation.

L.H.S. = (cos A/(1+sin A)) + ((1+sin A)/cos A)

= [cos 2 A + (1+sin A) 2 ]/(1+sin A)cos A

= (cos 2 A + sin 2 A + 1 + 2sin A)/(1+sin A) cos A

Since cos 2 A + sin 2 A = 1, we can write it as

= (1 + 1 + 2sin A)/(1+sin A) cos A

= (2+ 2sin A)/(1+sin A)cos A

= 2(1+sin A)/(1+sin A)cos A

= 2/cos A = 2 sec A = R.H.S.

L.H.S. = R.H.S.

(cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)

We know that tan θ =sin θ/cos θ

Now, substitute it in the given equation, to convert it in a simplified form

= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]

= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]

= sin 2 θ/[cos θ(sin θ-cos θ)] + cos 2 θ/[sin θ(cos θ-sin θ)]

= sin 2 θ/[cos θ(sin θ-cos θ)] – cos 2 θ/[sin θ(sin θ-cos θ)]

= 1/(sin θ-cos θ) [(sin 2 θ/cos θ) – (cos 2 θ/sin θ)]

= 1/(sin θ-cos θ) × [(sin 3 θ – cos 3 θ)/sin θ cos θ]

= [(sin θ-cos θ)(sin 2 θ+cos 2 θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]

= (1 + sin θ cos θ)/sin θ cos θ

= 1/sin θ cos θ + 1

= 1 + sec θ cosec θ = R.H.S.

Therefore, L.H.S. = R.H.S.

Hence proved

(iv)  (1 + sec A)/sec A = sin 2 A/(1-cos A)

First find the simplified form of L.H.S

L.H.S. = (1 + sec A)/sec A

Since secant function is the inverse function of cos function and it is written as

= (1 + 1/cos A)/1/cos A

= (cos A + 1)/cos A/1/cos A

Therefore, (1 + sec A)/sec A = cos A + 1

R.H.S. = sin 2 A/(1-cos A)

We know that sin 2 A = (1 – cos 2 A), we get

= (1 – cos 2 A)/(1-cos A)

= (1-cos A)(1+cos A)/(1-cos A)

Therefore, sin 2 A/(1-cos A)= cos A + 1

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A, using the identity cosec 2 A = 1+cot 2 A.

With the help of identity function, cosec 2 A = 1+cot 2 A, let us prove the above equation.

L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)

Divide the numerator and denominator by sin A, we get

= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A

We know that cos A/sin A = cot A and 1/sin A = cosec A

= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)

= (cot A – cosec 2 A + cot 2 A + cosec A)/(cot A+ 1 – cosec A) (using cosec 2 A – cot 2 A = 1

= [(cot A + cosec A) – (cosec 2 A – cot 2 A)]/(cot A+ 1 – cosec A)

= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)

=  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)

=  cot A + cosec A = R.H.S.

Therefore, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A

Hence Proved

First divide the numerator and denominator of L.H.S. by cos A,

We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes,

= √(sec A+ tan A)/(sec A-tan A)

Now using rationalization, we get

= (sec A + tan A)/1

= sec A + tan A = R.H.S

(vii) (sin θ – 2sin 3 θ)/(2cos 3 θ-cos θ) = tan θ

L.H.S. = (sin θ – 2sin 3 θ)/(2cos 3 θ – cos θ)

Take sin θ as in numerator and cos θ in denominator as outside, it becomes

= [sin θ(1 – 2sin 2 θ)]/[cos θ(2cos 2 θ- 1)]

We know that sin 2 θ = 1-cos 2 θ

= sin θ[1 – 2(1-cos 2 θ)]/[cos θ(2cos 2 θ -1)]

= [sin θ(2cos 2 θ -1)]/[cos θ(2cos 2 θ -1)]

= tan θ = R.H.S.

(viii) (sin A + cosec A) 2  + (cos A + sec A) 2  = 7+tan 2 A+cot 2 A

L.H.S. = (sin A + cosec A) 2  + (cos A + sec A) 2

It is of the form (a+b) 2 , expand it

(a+b) 2 =a 2 + b 2 +2ab

= (sin 2 A + cosec 2 A + 2 sin A cosec A) + (cos 2 A + sec 2 A + 2 cos A sec A)

= (sin 2 A + cos 2 A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan 2 A + 1 + cot 2 A

= 1 + 2 + 2 + 2 + tan 2 A + cot 2 A

= 7+tan 2 A+cot 2 A = R.H.S.

Therefore, (sin A + cosec A) 2  + (cos A + sec A) 2  = 7+tan 2 A+cot 2 A

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

First, find the simplified form of L.H.S

L.H.S. = (cosec A – sin A)(sec A – cos A)

Now, substitute the inverse and equivalent trigonometric ratio forms

= (1/sin A – sin A)(1/cos A – cos A)

= [(1-sin 2 A)/sin A][(1-cos 2 A)/cos A]

= (cos 2 A/sin A)×(sin 2 A/cos A)

= cos A sin A

Now, simplify the R.H.S

R.H.S. = 1/(tan A+cotA)

= 1/(sin A/cos A +cos A/sin A)

= 1/[(sin 2 A+cos 2 A)/sin A cos A]

(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

(x)  (1+tan 2 A/1+cot 2 A) = (1-tan A/1-cot A) 2  =   tan 2 A

L.H.S. = (1+tan 2 A/1+cot 2 A)

Since cot function is the inverse of tan function,

= (1+tan 2 A/1+1/tan 2 A)

= 1+tan 2 A/[(1+tan 2 A)/tan 2 A]

Now cancel the 1+tan 2 A terms, we get

(1+tan 2 A/1+cot 2 A) = tan 2 A

(1-tan A/1-cot A) 2  =   tan 2 A

NCERT Solutions for Class 10 Chapter 8 – Introduction to Trigonometry

For the  Class 10 CBSE Maths examinations, out of the 80 marks (combined), 12 marks are assigned from the unit 5 “Trigonometry”. The paper consists of 4 parts. Each part carries different marks and the questions have been assigned with 1 mark, two marks, 3 marks and 4 marks. You can expect at least 2-3 compulsory questions from this chapter. The main topics covered in this chapter include:

8.1 Introduction

You have already studied about triangles, and in particular, right triangles, in your earlier classes. In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. We also define the trigonometric ratios for angles of measure 0 o and 90 o . We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.

8.2 Trigonometric Identities

You have studied the concept of ratio, in your earlier classes. We now define certain ratios involving the sides of a right triangle, and call them trigonometric ratios. The topic is explained with suitable examples by using different functions of Trigonometry.

8.3 Trigonometric Ratios of Some Specific Angles

From Geometry, you are already familiar with the construction of angles of 30 0 , 45 o , 60 o and 90 0 . In this section, we will find the values of the trigonometric ratios for these angles and for 0 o . It explains Trigonometric Ratios of 45 o , Trigonometric Ratios of 30 o and 60 0 , Trigonometric Ratios of 0 o and 90 o with suitable examples.

8.4 Trigonometric Ratios of Complementary Angles

Two angles are said to be complementary if their sum equals 90 o . The topic discusses various formulas to solve numerical problems related to trigonometric ratios.

8.5 Trigonometric Identities

You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true of all values of the angles involved. In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities.

8.6 Summary

Summary is the brief of concepts that includes all the important points which you need to memorize to solve numerical problems related to the chapter.

List of Exercises in Class 10 Maths Chapter 8 :

Exercise 8.1 Solutions – 11 Questions (7 short answers, 3 long answers, 1 short answer with reasoning)

Exercise 8.2 Solutions – 4 Questions ( 1 short answer, 2 long answers, 1 MCQ)

Exercise 8.3 Solutions – 7 Questions (5 short answers, 2 long answers)

Exercise 8.4 Solutions – 5 Questions ( 2 short answers, 2 long answers, 1 MCQ)

Hence, these  NCERT Solutions for Class 10 Maths  will help students understand different types of questions and their answers along with key shortcuts and diagrammatic representations. All the NCERT Solutions for Class 10 Maths Chapter 8   PDF given here are presented in simple language. Comprehending these solutions thoroughly will aid students to solve complex problems effortlessly.

The faculty have curated the NCERT Solutions for Class 10 in a lucid manner to improve the problem-solving abilities among the students. For a more clear idea about Introduction To Trigonometry, students can refer to the study materials available at BYJU’S.

• RD Sharma Solutions for Class 10 Maths Introduction To Trigonometry

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I can’t understand how 6/4√3=√3/2

we have, 6/4√3=√3/2 6/4√3= 3/2√3 (√3 * √3)/2√3 Since,√3/√3 = 1 we get the answer, √3/2

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Class 10 Maths NCERT Solutions Chapter 8 Introduction to Trigonometry

• January 17, 2020 January 28, 2020

Class 10 Maths NCERT Solutions Chapter 8: All the Solutions prevailing here related to Class 10 Maths Chapter 8 will make your preparation easier. Understand the concepts and score well in the Board Examination with the notes, important questions, Multiple Choice Questions, Exemplar Problems listed for your convenience. NCERT Solutions for Class 10 Maths Introduction to Trigonometry are prepared by referring to the latest NCERT Textbooks in accordance with UP, MP and CBSE Boards.

Maths NCERT Solutions Class 10 Chapter 8 acts as a strong foundation for each and every concept. Clear your doubts and understand the fundamentals present by practicing Exercise Wise(8,1, 8.2, 8.3, 8.4) Solutions for NCERT Class 10 maths. Apart from these, you will get a direct link to download NCERT Maths Class 10 Introduction to Trigonometry Solutions PDF free of cost. You will have an idea on the Topics and Subtopics listed in Class 10 Maths Chapter 8 i.e. Introduction to Geometry. Students can make use of Chapter 8 Class 10 NCERT Maths Solutions in both Hindi and English Mediums on this page.

Topics and Subtopics Prevailing in Introduction to Geometry

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry

Solved answers for Class 10 Maths Chapter 8 will be helpful during your preparation for the board examination. Make sure you solve all the exercises in Introduction to Trigonometry so that you can score well. Our major concern is to provide step by step solutions for CBSE Class 10 Maths Chapter 8 in a way to simplify the complex concepts too. Brainstorming the formulas, concepts with a preparation strategy can yield you better results. You will have all the resources needed for your preparation like Study Material, Notes, Important Questions, etc.

Class 10 Maths NCERT Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Class 10 Maths NCERT Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Class 10 Maths NCERT Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Class 10 Maths NCERT Solutions Chapter 8 Introduction to Trigonometry Ex 8.4

Maths NCERT Solutions Class 10 Chapter 8 – Solved Exercises

Quality Solutions for CBSE Class 10 Maths are prepared by our experts by strictly following the guidelines as per the latest NCERT Textbooks and Study Materials revised from time to time. You can either use them online or download the NCERT Solutions for Class 10 Maths PDF and use it for further reference.

When it comes to Chapter 8 Introduction to Trigonometry the total weightage would be 8 Marks. There will be a 1 Mark, 3 Mark, and 4 Marks Question from Chapter 8 in the Board Examination. Exercise 8.4 of this chapter is important and there are chances of getting a 4 mark question from this so practice more of it.

It would be easy for you to solve even difficult questions after the rigorous practice of CBSE Class 10 Maths Solutions. Confidence will build up and you will attempt the exam well and score good marks if you practice on a regular basis. To make it easy for you we tried providing exercise 8.1 Class 10 Maths NCERT Solutions, ex 8.2 maths Class 10 Solutions, ex 8.3 maths class 10 solutions, exercise 8.4 Class 10 NCERT Solutions for Maths all in detail here.

We believe the information shared above regarding Class 10 Maths NCERT Solutions Chapter 8 as per our knowledge is helpful. In case of any queries feel free to drop a comment and share your feedback with us.

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NCERT Solutions for Class 10 Maths Ch 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry 

• Exercise 8.1
• Exercise 8.2
• Exercise 8.3
• Exercise 8.4

How many exercises in Chapter 8 Introduction to Trigonometry

If tan a = cot b, prove that a + b = 90°., what is the value of sinθ. cos(90° − θ) + cosθ . sin(90° − θ), express sin 67° + cos 75° in terms of ratios of angles between 0° and 45°., contact form.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

NCERT Solutions Class 10 Maths Chapter 8 Introduction To Trigonometry are provided here to help the students of CBSE class 10. Our expert teachers prepared all these solutions as per the latest CBSE syllabus and guidelines. CBSE Class 10 Maths solutions provide a detailed and step-wise explanation of each answer to the questions given in the exercises of NCERT books.

CBSE Class 10 Maths Chapter 8 Introduction To Trigonometry Solutions

Below we have given the answers to all the questions present in Introduction To Trigonometry in our NCERT Solutions for Class 10 Maths chapter 8. In this lesson, students are introduced to a lot of important concepts that will be useful for those who wish to pursue mathematics as a subject in their future classes. Based on these solutions, students can prepare for their upcoming Board Exams. These solutions are helpful as the syllabus covered here follows NCERT guidelines.

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4

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