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

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

Edited By Ramraj Saini | Updated on Dec 03, 2023 01:43 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 2 Exercise 2.1

NCERT Solutions for Exercise 2.1 Class 12 Maths Chapter 2 Inverse Trigonometric Functions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT Solutions for Class 12 Maths chapter 2 exercise 2.1 discusses the inverse of various trigonometric functions. From the Exercise 2.1 Class 12 Maths it can be observed that most of the questions are related to finding the value of inverse of various trigonometric functions like sine, cos, tan etc. From NCERT Solutions for Class 12 Maths chapter 2 exercise 2.1, direct questions related to finding the inverse are asked many times in board exams. Hence it is highly recommended to practice this exercise before CBSE class 12 board exam. Sometimes questions are asked in competitive exams also like JEE main etc. The NCERT chapter Inverse Trigonometric Functions has a lot of applications in subsequent chapters of maths as well as Physics also.

12th class Maths exercise 2.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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

Let $x = \sin^{-1}\left ( \frac{-1}{2} \right )$

$\implies \sin x = \frac{-1}{2}= -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$
We know, principle value range of $\dpi{100} sin^{-1}$ is $\dpi{100} [-\frac{\pi}{2}, \frac{\pi}{2}]$

$\dpi{100} \therefore$ The principal value of $\dpi{100} \sin^{-1}\left ( \frac{-1}{2} \right )$ is $\dpi{100} -\frac{\pi}{6},$

So, let us assume that $\cos^{-1}\left(\frac{\sqrt3}{2} \right ) = x$ then,

Taking inverse both sides we get;

$cos\ x = (\frac{\sqrt{3}}{2})$ , or $cos (\frac{\pi}{6}) = (\frac{\sqrt{3}}{2})$

and as we know that the principal values of $cos^{-1}$ is from [0, $\pi$ ],

Hence $cos\ x = (\frac{\sqrt{3}}{2})$ when x = $\frac{\pi}{6}$ .

Therefore, the principal value for $\cos^{-1}\left(\frac{\sqrt3}{2} \right )$ is $\frac{\pi}{6}$ .

Let us assume that $\textup{cosec}^{-1}(2) = x$ , then we have;

$Cosec\ x = 2$ , or

$Cosec( \frac{\pi}{6}) = 2$ .

And we know the range of principal values is $[\frac{-\pi}{2},\frac{\pi}{2}] - \left \{ 0 \right \}.$

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

Let us assume that $\tan^{-1}(-\sqrt3) = x$ , then we have;

$\tan x = (-\sqrt 3)$ or

$-\tan (\frac{\pi}{3}) = \tan \left ( \frac{-\pi}{3} \right ).$

and as we know that the principal value of $\tan^{-1}$ is $\left ( \frac{-\pi}{2}, \frac{\pi}{2} \right )$ .

Hence the only principal value of $\tan^{-1}(-\sqrt3)$ when $x = \frac{-\pi}{3}$ .

Let us assume that $\cos^{-1}\left(-\frac{1}{2} \right ) =y$ then,

Easily we have; $\cos y = \left ( \frac{-1}{2} \right )$ or we can write it as:

$-\cos \left ( \frac{\pi}{3} \right ) = \cos \left ( \pi - \frac{\pi}{3} \right ) = \cos \left ( \frac{2\pi}{3} \right ).$

as we know that the range of the principal values of $\cos^{-1}$ is $\left [ 0,\pi \right ]$ .

Hence $\frac{2\pi}{3}$ lies in the range it is a principal solution.

Given $\tan^{-1}(-1)$ so we can assume it to be equal to 'z';

$\tan^{-1}(-1) =z$ ,

$\tan z = -1$

or

$-\tan (\frac{\pi}{4}) = \tan(\frac{-\pi}{4})= -1$

And as we know the range of principal values of $\tan^{-1}$ from $\left ( \frac{-\pi}{2}, \frac{\pi}{2} \right )$ .

As only one value z = $-\frac{\pi}{4}$ lies hence we have only one principal value that is $-\frac{\pi}{4}$ .

Let us assume that $\sec^{-1}\left (\frac{2}{\sqrt3}\right) = z$ then,

we can also write it as; $\sec z = \left (\frac{2}{\sqrt3}\right)$ .

Or $\sec (\frac{\pi}{6}) = \left (\frac{2}{\sqrt3}\right)$ and the principal values lies between $\left [ 0, \pi \right ] - \left \{ \frac{\pi}{2} \right \}$ .

Hence we get only one principal value of $\sec^{-1}\left (\frac{2}{\sqrt3}\right)$ i.e., $\frac{\pi}{6}$ .

Let us assume that $\cot^{-1}(\sqrt3) = x$ , then we can write in other way,

$\cot x = (\sqrt3)$ or

$\cot (\frac{\pi}{6}) = (\sqrt3)$ .

Hence when $x=\frac{\pi}{6}$ we have $\cot (\frac{\pi}{6}) = (\sqrt3)$ .

and the range of principal values of $\cot^{-1}$ lies in $\left ( 0, \pi \right )$ .

Then the principal value of $\cot^{-1}(\sqrt3)$ is $\frac{\pi}{6}$

Let us assume $\cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x$ ;

Then we have $\cos x = \left ( \frac{-1}{\sqrt 2} \right )$

or

$-\cos (\frac{\pi}{4}) = \left ( \frac{-1}{\sqrt 2} \right )$ ,

$\cos (\pi - \frac{\pi}{4}) = \cos (\frac{3\pi}{4})$ .

And we know the range of principal values of $\cos^{-1}$ is $[0,\pi]$ .

So, the only principal value which satisfies $\cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x$ is $\frac{3\pi}{4}$ .

Let us assume the value of $\textup{cosec}^{-1}(-\sqrt2) = y$ , then

we have $cosec\ y = (-\sqrt 2)$ or

$-cosec\ (\frac{\pi}{4}) = (-\sqrt 2) = cosec\ (\frac{-\pi}{4})$ .

and the range of the principal values of $\textup{cosec}^{-1}$ lies between $\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ] - \left \{ 0 \right \}$ .

hence the principal value of $\textup{cosec}^{-1}(-\sqrt2)$ is $\frac{-\pi}{4}$ .

To find the values first we declare each term to some constant ;

$tan^{-1}(1) = x$ , So we have $\tan x = 1$ ;

or $\tan (\frac{\pi}{4}) = 1$

Therefore, $x = \frac{\pi}{4}$

$cos^{-1}(\frac{-1}{2}) = y$

So, we have

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

Therefore $y = \frac{2\pi}{3}$ ,

$\sin^{-1}(\frac{-1}{2}) = z$ ,

So we have;

$\sin z = \frac{-1}{2}$ or $-\sin (\frac{\pi}{6}) =\sin (\frac{-\pi}{6}) = \frac{-1}{2}$

Therefore $z = -\frac{\pi}{6}$

Hence we can calculate the sum:

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

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

Here we have $\cos^{-1}\left(\frac{1}{2} \right ) + 2\sin^{-1}\left(\frac{1}{2} \right )$

let us assume that the value of

$\cos^{-1}\left ( \frac{1}{2} \right ) = x, \:and\:\sin^{-1}\left(\frac{1}{2} \right ) = y$ ;

then we have to find out the value of x +2y.

Calculation of x :

$\Rightarrow \cos^{-1}\left ( \frac{1}{2} \right ) = x$

$\Rightarrow \cos x = \frac{1}{2}$

$\Rightarrow \cos \frac{\pi}{3} = \frac{1}{2}$ ,

Hence $x = \frac{\pi}{3}$ .

Calculation of y :

$\Rightarrow \sin^{-1}\left(\frac{1}{2} \right ) = y$

$\Rightarrow \sin y = \frac{1}{2}$

$\Rightarrow \sin \frac{\pi}{6} = \frac{1}{2}$ .

Hence $y = \frac{\pi}{6}$ .

The required sum will be = $\frac{\pi}{3}+2(\frac{\pi}{6}) = \frac{2\pi}{3}$ .

(A) $0\leq y \leq \pi$

(B) $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

(C) $0 < y < \pi$

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

Given if $\sin^{-1}x = y$ then,

As we know that the $\sin^{-1}$ can take values between $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ].$

Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$ .

Hence answer choice (B) is correct.

(A) $\pi$

(B) $-\frac{\pi}{3}$

(C) $\frac{\pi}{3}$

(D) $\frac{2\pi}{3}$

Let us assume the values of $\tan^{-1}(\sqrt3)$ be 'x' and $\sec^{-1}(-2)$ be 'y'.

Then we have;

$\tan^{-1}(\sqrt3) = x$ or $\tan x = \sqrt 3$ or $\tan \frac{\pi}{3} = \sqrt 3$ or

$x = \frac{\pi}{3}$ .

and $\sec^{-1}(-2) = y$ or $\sec y = -2$

or $-\sec (\frac{\pi}{3}) =\sec ({\pi - \frac{\pi}{3}}) = \sec{\frac{2\pi}{3}}$

$y = \frac{2\pi}{3}$

also, the ranges of the principal values of $\tan^{-1}$ and $\sec^{-1}$ are $(\frac{-\pi}{2},\frac{\pi}{2})$ . and

$[0,\pi] - \left \{ \frac{\pi}{2} \right \}$ respectively.

$\therefore$ we have then;

$\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$

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

More About NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1

The NCERT book for Class 12 Maths chapter Inverse Trigonometric Functions has a total of 3 exercises including miscellaneous exercise. Exercise 2.1 Class 12 Maths covers solutions to a total of 21 questions mostly based on finding the inverse value of various trigonometric functions. NCERT Solutions for Class 12 Maths chapter 2 exercise 2.1 is an authentic source to learn concepts related to finding the inverse of a trigonometric function.

Benefits of NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1

• The NCERT syllabus Class 12th Maths chapter 2 exercise provided here is in detail which is solved by subject matter experts .
• Students are recommended to practice Exercise 2.1 Class 12 Maths to prepare for exams, direct questions are asked in Board exams as well as competitive exams.
• These Class 12 maths chapter 2 exercise 2.1 solutions can be referred by students to revise just before the exam for revision.
• NCERT Solutions for Class 12 Maths chapter 2 exercise 2.1 can be used to understand physics topics on similar concepts.
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## Key Features Of NCERT Solutions for Exercise 2.1 Class 12 Maths Chapter 2

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 2.1 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 2.1, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 2.1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 2.1 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 2.1 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 2.1 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

### Also see-

NCERT Solutions Subject Wise

## Subject wise NCERT Exemplar solutions

Happy learning!!!

### Frequently Asked Question (FAQs)

1. Which concepts are covered in Exercise 2.1 Class 12 Maths?

Concepts related finding the inverse of various trigonometric functions are covered in  Exercise 2.1 Class 12 Maths. More questions on Inverse Trigonometry can be solved using NCERT exemplar. Practice class 12 ex 2.1 to get deeper understanding of the concepts.

2. Mention important topics of class 12 maths chapter 2 Inverse Trigonometric Functions?

In ex 2.1 class 12, topics like finding the inverse of sine, cos, tan etc. are discussed that are asked frequently in the exam. Follow the NCERT syllabus to get a good score in the CBSE board exams.

3. How are the NCERT solutions helpful in the board exam ?

Most of the questions are asked directly from NCERT exercises in the Board examination. Hence it is advisable to go through the NCERT exercise.

4. Mention the total number of exercises in NCERT Class 12 Maths chapter 2 Inverse Trigonometric Functions?

In  NCERT class 12 maths chapter 2 Inverse Trigonometric Functions, there are a total of 3 exercises which includes a miscellaneous exercise also.

5. How many questions are covered in Exercise 2.1 Class 12 Maths ?

There are 14 questions in Exercise 2.1 Class 12 Maths

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

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Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

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Hello student,

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• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

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