NCERT Solutions for Exercise 2.1 Class 12 Maths Chapter 2

Q: 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.

Q: 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.

Q: 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.

Q: 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.

Q: How many questions are covered in Exercise 2.1 Class 12 Maths ?

There are 14 questions in Exercise 2.1 Class 12 Maths

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

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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.

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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

Question:1 Find the principal values of the following : $\sin^{-1}\left ( \frac{-1}{2} \right )$

Answer:

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 $sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$

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

Question:2 Find the principal values of the following: $\cos^{-1}\left(\frac{\sqrt3}{2} \right )$

Answer:

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 $\frac{\pi}{6}$ .

Question:3 Find the principal values of the following: $\textup{cosec}^{-1}(2)$

Answer:

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 $\frac{\pi}{6}$ .

Question:4 Find the principal values of the following: $\tan^{-1}(-\sqrt3)$

Answer:

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

Question:5 Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{2} \right )$

Answer:

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.

Question:6 Find the principal values of the following : $\tan^{-1}(-1)$

Answer:

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

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

$\tan z = -1$

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

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 $-\frac{\pi}{4}$ .

Question:7 Find the principal values of the following : $\sec^{-1}\left (\frac{2}{\sqrt3}\right)$

Answer:

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., $\frac{\pi}{6}$ .

Question:8 Find the principal values of the following: $\cot^{-1}(\sqrt3)$

Answer:

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 $\frac{\pi}{6}$

Question:9 Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{\sqrt2} \right )$

Answer:

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

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

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

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

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 $\frac{3\pi}{4}$ .

Question:10 Find the principal values of the following: $\textup{cosec}^{-1}(-\sqrt2)$

Answer:

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 $\frac{-\pi}{4}$ .

Question:11 Find the values of the following: $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2} \right ) + \sin^{-1}\left(-\frac{1}{2} \right )$

Answer:

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:

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

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

Question:12 Find the values of the following: $\cos^{-1}\left(\frac{1}{2} \right ) + 2\sin^{-1}\left(\frac{1}{2} \right )$

Answer:

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 = $\frac{\pi}{3}+2(\frac{\pi}{6}) = \frac{2\pi}{3}$ .

Question:13 If $\sin^{-1}x = y$ then

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

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

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

Answer:

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.

Question:14 $\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$ is equal to

(A) $\pi$

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

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

Answer:

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.

Also Read| Inverse Trigonometric Functions NCERT Notes

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.

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Happy learning!!!

Frequently Asked Questions (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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Prachi Kumari 30 December,2024

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

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.

hope this helps.

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Ashvadha 12 July,2024

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Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
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Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

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

NCERT Solutions For Class 12 Maths Chapter 2 Exercise 2.1

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

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