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

## NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Two chapters 'relation and function' and 'inverse trigonometry' of NCERT Class 12 has 10 % weightage in the board examination. In this article, you will find NCERT solutions for class 12 maths chapter 2 inverse trigonometric functions which will help you to understand the concepts in a detailed manner. In class 11 maths you have already learnt about trigonometric functions. It won't take much effort to command on inverse trigonometric functions if you have good knowledge of trigonometric functions. You just need to practice NCERT questions including examples and miscellaneous exercise. You may find some difficulties in solving the problems, so you can take the help of these solutions of NCERT for class 12 maths chapter 2 inverse trigonometric functions. These solutions of NCERT for class 12 maths chapter 2 inverse trigonometric functions are very important for the board exam and as well as for competitive exams like JEE Main, BITSAT, VITEEE, etc. In this chapter, there are 2 exercises with 35 questions. All these questions are prepared and explained in a step-by-step method in the NCERT solutions for class 12 maths chapter 2 inverse trigonometric functions article.  So, it will be very easy for you to understand the concept. Check all  NCERT solutions from class 6 to class 12 to get a better understanding of the concepts.

There are important applications of ITF in geometry, navigation, science, and engineering. Also, inverse trigonometric functions play an important role in the calculus part of mathematics to define many integrals. Many students have a misconception in class 12 maths chapter inverse trigonometric functions like  . But the inverse function () is not the same as   for example .

For inverse to exist, the function must be one-one and onto but trigonometric functions are neither one-one and onto over their domain and natural ranges. So, to ensure the existence of their inverse we restrict domains and ranges of trigonometric functions. And this range is known as principal value. In the following table the principal value branches of inverse trigonometric functions(ITF) are given: ## Topics of NCERT Grade 12 Maths Chapter-2 Inverse Trigonometric Functions

2.1 Introduction

2.2 Basic Concepts

2.3 Properties of Inverse Trigonometric Functions

## CBSE NCERT Solutions for class 12 maths chapter 2 Inverse Trigonometric Functions: Exercise 2.1

Let

We know, principle value range of  is

The principal value of  is

So, let us assume that   then,

Taking inverse both sides we get;

,  or

and as we know that the principal values of  is from [0,],

Hence      when x =  .

Therefore, the principal value for is   .

Let us assume that , then we have;

, or

.

And we know the range of principal values is

Therefore the principal value of   is .

Let us assume that , then we have;

or

and as we know that the principal value of  is .

Hence the only principal value of  when .

Let us assume that   then,

Easily we have;  or we can write it as:

as we know that the range of the principal values of   is .

Hence  lies in the range it is a principal solution.

Question:6 Find the principal values of the following:

Given  so we can assume it to be equal to 'z';

,

or

And as we know the range of principal values of  from .

As only one value z =  lies hence we have only one principal value that is .

Question:7 Find the principal values of the following:

Let us assume that   then,

we can also write it as; .

Or   and the principal values lies between  .

Hence we get only one principal value of   i.e., .

Let us assume that  , then we can write in other way,

or

.

Hence when     we have .

and the range of principal values of  lies in .

Then the principal value of  is

Let us assume ;

Then we have

or

,

.

And we know the range of principal values of  is .

So, the only principal value which satisfies  is .

Question:10 Find the principal values of the following:

Let us assume the value of  , then

we have    or

.

and the range of the principal values of   lies between  .

hence the principal value of  is .

Question:11 Find the values of the following:

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

, So we have ;

or

Therefore,

So, we have

.

Therefore ,

,

So we have;

or

Therefore

Hence we can calculate the sum:

.

Question:12 Find the values of the following:

Here we have

let us assume that the value of

;

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

Calculation of x :

,

Hence .

Calculation of y :

.

Hence .

The required sum will be =     .

Question:13 If  then

(A)

(B)

(C)

(D)

Given if    then,

As we know that the  can take values between

Therefore, .

Hence answer choice (B) is correct.

Question:14   is equal to

(A)

(B)

(C)

(D)

Let us assume the values of  be 'x'  and   be 'y'.

Then we have;

or       or     or

and      or

or

also, the ranges of the principal values of  and     are . and

respectively.

we have then;

## CBSE NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions: Exercise 2.2

Question:1 Prove the following:

Given to prove:

where,     .

Take   or

Take R.H.S value

=   L.H.S

Question:2 Prove the following:

Given to prove .

Take   or ;

Then we have;

R.H.S.

= L.H.S

Hence Proved.

Question:3 Prove the following:

Given to prove

We have L.H.S

=   R.H.S

Hence proved.

Question:4 Prove the following:

Given to prove

Then taking L.H.S.

We have

=  R.H.S.

Hence proved.

We have

Take

is the simplified form.

Given that

Take   or

=

Given that

We have in inside the root the term :

Put    and    ,

Then we have,

Hence the simplest form is

Given    where

So,

Taking  common from numerator and denominator.

We get:

as,

is the simplest form.

Given that

Take    or

and putting it in the equation above;

is the simplest form.

Given

Here we can take

So,

will become;

and as ;

hence the simplest form is .

Question:11 Find the values of each of the following:

Given equation:

So, solving the inner bracket first, we take the value of

Then we have,

Therefore, we can write .

.

Question:12 Find the values of each of the following:

We have to find the value of

As we know    so,

Equation reduces to .

Question:13 Find the values of each of the following:  and

Taking the value   or    and   or  then we have,

,

Then,

Ans.

Question:14  If , then find the value of .

As we know the identity;

. it will just hit you by practice to apply this.

So,      or    ,

we can then write ,

putting in above equation we get;

Ans.

Question:15 If , then find the value of .

Using the identity ,

We can find the value of x;

So,

on applying,

or   ,

Hence, the possible values of x are .

Given ;

We know that

If the value of x belongs to   then we get the principal values of .

Here,

We can write   is as:

=   where

As we know

If  which is the principal value range of .

So, as in ;

Hence we can write    as :

=

Where

and

Given that

we can take ,

then

or

We have similarly;

Therefore we can write

from

Question:19  is equal to

(A)

(B)

(C)

(D)

As we know that  if  and is principal value range of .

In this case ,

hence we have then,

Hence the correct answer is  (B).

Question:20  is equal to

(A)

(B)

(C)

(D)

Solving the inner bracket of ;

or

Take   then,

and we know the range of principal value of

Therefore we have .

Hence,

Hence the correct answer is D.

Question:21   is equal to

(A)

(B)

(C)    0

(D)

We have ;

finding the value of  :

Assume  then,

and the range of the principal value of  is .

Hence, principal value is

Therefore

and

so, we have now,

or,

Hence the answer is option  (B).

### Solutions of NCERT for class 12 maths chapter 2 Inverse Trigonometric Functions: Miscellaneous Exercise

Question:1 Find the value of the following:

If   then   , which is principal value of .

So, we have

Therefore we have,

.

Question:2 Find the value of the following:

We have given ;

so, as we know

So, here we have .

Therefore we can write  as:

.

Question:3 Prove that

To prove: ;

Assume that

then we have .

or

Therefore we have

Now,

We can write L.H.S as

as we know

L.H.S = R.H.S

Question:4 Prove that

Taking

then,

Therefore we have-

.............(1).

,

Then,

.............(2).

So, we have now,

L.H.S.

using equations (1) and (2) we get,

=  R.H.S.

Question:5 Prove that

Take   and     and

then we have,

Then we can write it as:

or

...............(1)

Now,

So,                          ...................(2)

Also we have similarly;

Then,

...........................(3)

Now, we have

L.H.S

so, using (1) and (2) we get,

or we can write it as;

=  R.H.S.

Hence proved.

Question:6 Prove that

Converting all terms in tan form;

Let  ,    and .

now, converting all the terms:

or

We can write it in tan form as:

.

or            ................(1)

or

We can write it in tan form as:

or          ......................(2)

Similarly, for ;

we have      .............(3)

Using (1) and (2) we have L.H.S

On applying

We have,

...........[Using (3)]

=R.H.S.

Hence proved.

Question:7 Prove that

Taking R.H.S;

We have

Converting sin and cos terms in tan forms:

Let    and

now, we have    or

............(1)

Now,

................(2)

Now, Using (1) and (2) we get,

R.H.S.

as we know

so,

equal to L.H.S

Hence proved.

Question:8 Prove that

Applying the formlua:

on two parts.

we will have,

Hence it s equal to R.H.S

Proved.

Question:9 Prove that

By observing the square root we will first put

.

Then,

we have

or, R.H.S.

.

L.H.S.

hence L.H.S. = R.H.S proved.

Question:10 Prove that

Given that

By observing we can rationalize the fraction

We get then,

Therefore we can write it as;

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

Hence proved.

Question:11 Prove that

[Hint: Put ]

By using the Hint we will put ;

we get then,

dividing numerator and denominator by ,

we get,

using the formula

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

Hence proved

Question:12 Prove that

We have to solve the given equation:

Take as common in L.H.S,

or     from

Now, assume,

Then,

Therefore we have now,

So we have L.H.S then

That is equal to R.H.S.

Hence proved.

Question:13 Solve the following equations:

Given equation ;

Using the formula:

We can write

So, we can equate;

that implies that .

or          or

Hence we have solution .

Question:14 Solve the following equations:

Given equation is

:

L.H.S can be written as;

Using the formula

So, we have

Hence the value of .

Question:15   is equal to

(A)

(B)

(C)

(D)

Let then we have;

or

Hence the correct answer is D.

Question:16  then  is equal to

(A)

(B)

(C)    0

(D)

Given the equation:

we can migrate the  term to the R.H.S.

then we have;

or                                ............................(1)

from

Take       or    .

So, we conclude that;

Therefore we can put the value of  in equation (1)  we get,

Putting x= sin y, in the above equation; we have then,

So, we have the solution;

Therefore we have .

When we have , we can see that :

So, it is not equal to the R.H.S.

Thus we have only one solution which is x = 0

Hence the correct answer is (C).

Question:17 is equal to

(A)

(B)

(C)

(D)

Applying formula: .

We get,

Hence, the correct answer is C.

## NCERT solutions for class 12 maths chapter-wise

 chapter 1 Solutions of NCERT for class 12 maths chapter 1 Relations and Functions chapter 2 NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions chapter 3 NCERT solutions for class 12 maths chapter 3 Matrices chapter 4 CBSE NCERT solutions for class 12 maths chapter 4 Determinants chapter 5 Solutions of NCERT for class 12 maths chapter 5 Continuity and Differentiability chapter 6 CBSE NCERT solutions for class 12 maths chapter 6 Application of Derivatives chapter 7 NCERT solutions for class 12 maths chapter 7 Integrals chapter 8 Solutions of NCERT for class 12 maths chapter 8 Application of Integrals chapter 9 CBSE NCERT solutions for class 12 maths chapter 9 Differential Equations chapter 10 NCERT solutions for class 12 maths chapter 10 Vector Algebra chapter 11 Solutions of NCERT for class 12 maths chapter 11 Three Dimensional Geometry chapter 12 CBSE NCERT solutions for class 12 maths chapter 12 Linear Programming chapter 13 NCERT solutions for class 12 maths chapter 13 Probability

## NCERT solutions for class 12

Students who wish to perform well in the CBSE 12 board examination, solutions of NCERT for class 12 maths chapter 2 inverse trigonometric functions are very helpful but here are some tips to make command on Inverse Trigonometric Functions.

• The inverse trigonometric function is inverse of the trigonometric function, so if you have a command on the trigonometric function then it will be easy for you to understand inverse trigonometric functions.
• Try to relate trigonometric functions formulas with inverse trigonometric functions formulas, so that memorizing the formulae becomes easier.
• Before starting to solve exercise, first solve the examples that are given in the NCERT class 12 maths textbook.
• Also, try to solve every exercise including miscellaneous exercise, NCERT chapter examples, miscellaneous examples on your own, if you are finding difficulties, you can take the help of CBSE NCERT solutions for class 12 maths chapter 2 inverse trigonometric functions.
• If you have solved all NCERT then you can solve previous years paper CBSE board to get familiar with the pattern of board exam question paper

Happy learning!!!