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# NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1 - Inverse Trigonometric Functions

Last updated date: 11th Sep 2024
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## NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1 - FREE PDF Download

NCERT Solutions for Maths Chapter 2 Exercise 2.1 for Class 12, provided by Vedantu, focus on introducing inverse trigonometric functions. This exercise covers the definitions, domains, ranges, and principal values of Trigonometric Functions. It helps students understand how to find the specific values (principal values) within a given range for these functions, and learn how to draw graphs of Trigonometric Functions. Ex 2.1 Class 12 includes various problems that require applying the definitions and properties of inverse trigonometric functions. Vedantu’s step-by-step solutions ensure students build a strong foundation and can solve related problems confidently. You can download the FREE PDF for NCERT Solutions for Class 12 Maths from Vedantu’s website and boost your preparations for Exams.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1 - FREE PDF Download
2. Glance on NCERT Solutions  Class 12 Maths Chapter 2 Exercise 2.1 | Vedantu
3. Important Formulas Used in Class 12 Chapter 2 Exercise 2.1
4. Access NCERT Solutions for Maths Class 12 Chapter 2 - Trigonometric Functions Exercise 2.1
5. Class 12 Maths Chapter Inverse Trigonometric Functions : Exercises Breakdown
6. CBSE Class 12 Maths Chapter 2 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

## Glance on NCERT Solutions  Class 12 Maths Chapter 2 Exercise 2.1 | Vedantu

• Class 12 Ex 2.1 of Chapter 3 focuses on domains, ranges, and principal values of Trigonometric Functions.

• Introduction to Inverse Trigonometric Functions: Covers functions like Sin-1 x ,Cos-1 x,Tan-1 x,Cosec-1 x,Sec-1 x,Cot-1 x.

• Key Concepts covered in this chapter is to identify the unique values within specific ranges where these functions are defined.

• Identifying the domain and range for each inverse trigonometric function.

• Calculating the principal values for inverse trigonometric functions.

• There are links to video tutorials explaining class 12 chapter 2 Exercise 2.1 - Inverse Trigonometric Functions for better understanding.

• There are two examples and 14 questions covered in NCERT Class 12th Maths Chapter 2 Exercise 2.1 Inverse Trigonometric Functions.

## Important Formulas Used in Class 12 Chapter 2 Exercise 2.1

Some of the important theorems/properties of inverse trigonometric functions are given below. These theorems/properties help to convert one function into another.

 Function Domain Range Sin-1 x $\mathrm{\left [ -1,1 \right ]}$ $\mathrm{\left \{ -\frac{\pi }{2}, \frac{\pi }{2}\right \}}$ Cos-1 x $\mathrm{\left [ -1,1 \right ]}$ $\mathrm{\left \{ 0,\pi \right \}}$ Tan-1 x R $\mathrm{\left \{ -\frac{\pi }{2}, \frac{\pi }{2}\right \}}$ Cosec-1 x $\mathrm{x\geq 1\: or\: x\leq -1}$ $\mathrm{[0,\frac{\pi }{2})(\frac{\pi }{2},\pi ]}$ Sec-1 x $\mathrm{x\geq 1\: or\: x\leq -1}$ $\mathrm{[0,\frac{\pi }{2})(\frac{\pi }{2},\pi ]}$ Cot-1 x R $\mathrm{\left \{ 0,\pi \right \}}$

### Complementary Functions Related to Inverse Trigonometric Function

1. Sin-1 x + Cos-1 x = $\mathrm{\frac{\pi }{2}}$

2. Cosec-1 x + Sec-1 x = $\mathrm{\frac{\pi }{2}}$

3. Tan-1 x + Cot-1 x =  $\mathrm{\frac{\pi }{2}}$

4. Tan-1 x + Tan-1 y = $\mathrm{\tan ^{-1}\frac{x+y}{1-xy}\: xy< 1}$

5. Tan-1 x - Tan-1 y = $\mathrm{\tan ^{-1}\frac{x-y}{1+xy}\: xy< -1}$

Competitive Exams after 12th Science

## Access NCERT Solutions for Maths Class 12 Chapter 2 - Trigonometric Functions Exercise 2.1

1.  Find the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

Ans: Let’s assume when  ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y$, Then $\sin y = \left( { - \dfrac{1}{2}} \right) = - \sin \left( {\dfrac{\pi }{6}} \right) = \sin \left( { - \dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ and $\sin \left( { - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}$

Hence, the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $- \dfrac{\pi }{6}$.

2. Find the principal value of ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$

Ans: Let’s consider, ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = y.$ Then $\cos y = \dfrac{{\sqrt 3 }}{2} = \cos \left( {\dfrac{\pi }{6}} \right)$

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}$

Therefore, the principal value of ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ is $\dfrac{\pi }{6}$.

3. Find the principal value of ${\operatorname{cosec} ^{ - 1}}(2)$

Ans: Let’s consider,${\operatorname{cosec} ^{ - 1}}(2) = y$.Then, $\operatorname{cosec} {\text{y}} = 2 = \operatorname{cosec} \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\operatorname{cosec} ^{ - 1}}$ is$\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] - \{ 0\}$.

Therefore, the principal value of ${\operatorname{cosec} ^{ - 1}}(2)$ is $\dfrac{\pi }{6}$.

4. Find the principal value of ${\tan ^{ - 1}}( - \sqrt 3 )$

Ans: Let’s consider ${\tan ^{ - 1}}( - \sqrt 3 ) = y$ Then, $\tan y = - \sqrt 3 = - \tan \dfrac{\pi }{3} = \tan \left( { - \dfrac{\pi }{3}} \right)$.

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and$\tan \left( { - \dfrac{\pi }{3}} \right)$ is $- \sqrt 3$

Hence, the principal value of ${\tan ^{ - 1}}( - \sqrt 3 )$ is $- \dfrac{\pi }{3}$.

5. Find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

Ans: Let’s consider, ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y.$

Then, $\cos y = - \dfrac{1}{2} = - \cos \left( {\dfrac{\pi }{3}} \right) = \cos \left( {\pi - \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)$.

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{{2\pi }}{3}} \right) = - \dfrac{1}{2}$

Therefore, the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $\left( {\dfrac{{2\pi }}{3}} \right)$.

6. Find the principal value of ${\tan ^{ - 1}}( - 1)$

Ans: Let’s assume that ${\tan ^{ - 1}}( - 1) = {\text{y}}$.

Then, $\tan y = - 1 = - \tan \left( {\dfrac{\pi }{4}} \right) = \tan \left( { - \dfrac{\pi }{4}} \right)$.

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and$\tan \left( { - \dfrac{\pi }{4}} \right) = - 1$

Therefore, the principal value of ${\tan ^{ - 1}}( - 1)$ is $- \dfrac{\pi }{4}$.

7. Find the principal value of ${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)$

Ans: Let’s consider ${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right) = y$.

Then, $\sec y = \dfrac{2}{{\sqrt 3 }} = \sec \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\sec ^{ - 1}}$ is $[0,\pi ] - \left\{ {\dfrac{\pi }{2}} \right\}$ and $\sec \left( {\dfrac{\pi }{6}} \right) = \dfrac{2}{{\sqrt 3 }}$

8. Find the principal value of ${\cot ^{ - 1}}(\sqrt 3 )$

Ans: Let’s consider ${\cot ^{ - 1}}(\sqrt 3 ) = y$. Then $\cot y = \sqrt 3 = \cot \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\cot ^{ - 1}}$ is $(0,\pi )$ and $\cot \left( {\dfrac{\pi }{6}} \right) = \sqrt 3$.

Hence, the principal value of ${\cot ^{ - 1}}(\sqrt 3 )$ is $\dfrac{\pi }{6}$.

9. Find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$

Ans: Let’s ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = y$.

Then $\cos y = - \dfrac{1}{{\sqrt 2 }} = - \cos \left( {\dfrac{\pi }{4}} \right) = \cos \left( {\pi - \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{{3\pi }}{4}} \right)$.

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{{3\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$.

Therefore, the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$ is $\dfrac{{3\pi }}{4}$.

10. Find the principal value of ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 )$

Ans: Let’s consider, ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 ) = y$. Then, $\cos ecy = - \sqrt 2 = - \cos ec\left( {\dfrac{\pi }{4}} \right) = \cos ec\left( { - \dfrac{\pi }{4}} \right)$

As we know that the range of the principal value branch of ${\operatorname{cosec} ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] - \{ 0\}$ and $\operatorname{cosec} \left( { - \dfrac{\pi }{4}} \right) = - \sqrt 2$

Therefore, the principal value of ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 )$ is $- \dfrac{\pi }{4}$.

11. Find the value of ${\tan ^{ - 1}}(1) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

Ans: Let’s consider ${\tan ^{ - 1}}(1) = x$. Then, $\tan x = 1 = \tan \left( {\dfrac{\pi }{4}} \right)$.

$\therefore {\tan ^{ - 1}}(1) = \dfrac{\pi }{4}$

Let’s assume,${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y$.

Then, $\cos y = - \dfrac{1}{2} = - \cos \left( {\dfrac{\pi }{3}} \right) = \cos \left( {\pi - \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)$.

$\therefore {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{2\pi }}{3}$

Let’s again assume that ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = z$.

Then, $\sin z = - \dfrac{1}{2} = - \sin \left( {\dfrac{\pi }{6}} \right) = \sin \left( { - \dfrac{\pi }{6}} \right)$.

$\therefore {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = - \dfrac{\pi }{6}$

$\therefore {\tan ^{ - 1}}(1) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

$= \dfrac{\pi }{4} + \dfrac{{2\pi }}{3} - \dfrac{\pi }{6}$

$= \dfrac{{3\pi + 8\pi - 2\pi }}{{12}} = \dfrac{{9\pi }}{{12}} = \dfrac{{3\pi }}{4}$

12. Find the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$

Ans: Let’s consider,${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = x$.

Then, $\cos x = \dfrac{1}{2} = \cos \left( {\dfrac{\pi }{3}} \right)$.

$\therefore {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3}$

Let’s assume ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = y.$

Then, $\sin y = \dfrac{1}{2} = \sin \left( {\dfrac{\pi }{6}} \right)$.

$\therefore {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}$

$\therefore {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3} + 2 \times \dfrac{\pi }{6} = \dfrac{\pi }{3} + \dfrac{\pi }{3} = \dfrac{{2\pi }}{3}$

13. If ${\sin ^{ - 1}}x = y$, then

(A) $0 \leqslant {\text{y}} \leqslant \pi$

(B) $- \dfrac{\pi }{2} \leqslant y \leqslant \dfrac{\pi }{2}$

(C) $0 < y < \pi$

(D) $- \dfrac{\pi }{2} < y < \dfrac{\pi }{2}$

Ans: It is given that ${\sin ^{ - 1}}x = y$. As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$.

Therefore, $- \dfrac{\pi }{2} \leqslant y \leqslant \dfrac{\pi }{2}$.

Hence option (B) is correct.

14. ${\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}( - 2)$ is equal to

(A) $\pi$

(B) $- \pi /3$

(C) $\pi /3$

(D) $2\pi /3$

Ans: Let’s consider, ${\tan ^{ - 1}}\sqrt 3 = x.$.

Then, $\tan x = \sqrt 3 = \tan \dfrac{\pi }{3}$

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$. $\therefore {\tan ^{ - 1}}\sqrt 3 = \dfrac{\pi }{3}$

Let assume, ${\sec ^{ - 1}}( - 2) = y$.

Then, $\sec y = - 2 = - \sec \left( {\dfrac{\pi }{3}} \right) = \sec \left( {\pi - \dfrac{\pi }{3}} \right) = \sec \dfrac{{2\pi }}{3}$.

As we know that the range of the principal value branch of sec $^1$ is $[0,\pi ] - \left\{ {\dfrac{\pi }{2}} \right\}$. $\therefore {\sec ^{ - 1}}( - 2) = \dfrac{{2\pi }}{3}$

Thus, ${\tan ^{ - 1}}(\sqrt 3 ) - {\sec ^{ - 1}}( - 2)$

$= \dfrac{\pi }{3} - \dfrac{{2\pi }}{3} = - \dfrac{\pi }{3}$

## Conclusion

NCERT Solutions for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions by Vedantu offer a detailed guide to mastering key topics such as definitions, domains, ranges, principal values, and properties. Focus on understanding principal values and practicing properties to simplify expressions. These solutions provide a clear explanation of each problem, aiding in conceptual clarity. Vedantu ensures that students develop a solid grasp of the subject matter, making it easier to tackle complex problems. Additionally, the solutions are designed to align with the latest syllabus, ensuring thorough preparation for exams.

## Class 12 Maths Chapter Inverse Trigonometric Functions : Exercises Breakdown

 Exercise Number of Questions Exercise 2.2 20 Questions & Solutions (4 Short Answers, 16 Long Answers)

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1 - Inverse Trigonometric Functions

1. Which online platform offers NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.1?

Vedantu, the best online learning platform, provides authentic NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.1. The free PDF of exercise-wise NCERT Solutions for Chapter 2 Inverse Trigonometric Functions are designed by expert Maths tutors. The material is designed to help students revise the complete syllabus and score well in exams. NCERT Solutions for Class 12 Maths Chapter 2 are prepared as per the latest CBSE guidelines and exam pattern. It provides students with a thorough knowledge of the chapter and simple explanations to the exercise problems designed to help students in doubt clearance.

2. What are the learning objectives of Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.1?

In Exercise 2.1 of Class 12 Maths Chapter 2 Inverse Trigonometric Functions, students will learn about the inverse functions of the trigonometric functions. Students can get the solutions designed by experts for concept clarity. Students will be able to solve the Exercise 2.1 questions confidently without any mistakes after referring to NCERT Solutions for Class 12 Maths Chapter 2. Exercise 2.1 of Class 12 Maths Chapter 2 is crucial to develop the basics of Inverse Trigonometric Functions. The solutions are available in the free PDF format on Vedantu.

3. What are the benefits of exercise-wise NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions?

Class 12 Maths Chapter 2 Inverse Trigonometric Functions included in CBSE syllabus is an important chapter. The chapter explains the fundamental concepts related to inverse functions of trigonometric functions. NCERT Solutions for Class 12 Maths Chapter 2 are the most reliable resource available online. These solutions are provided by experts to clear all the doubts of students regarding this chapter. These experts have years of teaching experience. NCERT Solutions for Class 12 Chapter 2 Inverse Trigonometric Functions for Exercise 2.1 and other exercises is an excellent material to be a top scorer in the Maths exam.

4. What are the key aspects of NCERT Solutions by Vedantu for Class 12 Maths Chapter 2 Exercise 2.1 or any exercise?

NCERT Solutions for Class 12 Maths for Chapter 2 Inverse Trigonometric Functions are available on Vedantu’s platform (mobile app and website). Following are some of the key features of Vedantu’s NCERT Solutions for Class 12 Maths Chapter 2:

• Step-by-step explanations of solutions.

• Designed by subject matter experts with years of teaching experience.

• Provides complete coverage of the NCERT questions.

• Designed as per the latest syllabus and the exam pattern.

5. What topics are covered in exercise 2.1 class 12 maths solutions?

Exercise 2.1 covers the introduction to inverse trigonometric functions, their domains, ranges, and principal values.

6. How do you find the principal value of inverse trigonometric functions in class 12 maths chapter 2 exercise 2.1?

The principal value is found by determining the unique value of the inverse function within its principal branch.

7. What is the importance of understanding the domain and range of inverse trigonometric functions in class 12 maths ex 2.1?

Understanding the domain and range is crucial because it helps in solving equations and inequalities involving inverse trigonometric functions correctly.

8. What are the common mistakes to avoid while solving exercise 2.1 class 12 maths problems?

Common mistakes include using incorrect principal values, not considering the restricted domains, and confusing the inverse functions with their reciprocal trigonometric functions.

9. How can I effectively prepare for questions from 2.1 class 12 in exams?

Practice various problems from Exercise 2.1, understand the properties of inverse trigonometric functions, and review previous years' exam questions to get familiar with the types of questions asked.

10. Are there any real-life applications of the concepts learned in exercise 2.1 class 12?

Yes, inverse trigonometric functions are used in fields such as engineering, physics, and computer science for solving problems involving angles and distances.