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 stepbystep 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. Here you will get NCERT solutions for two exercises & a miscellaneous exercise of this chapter.
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 oneone and onto but trigonometric functions are neither oneone 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 Chapter2 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
Question:1 Find the principal values of the following :
Answer:
Let
We know, principle value range of
is
The principal value of is
Question:2 Find the principal values of the following:
Answer:
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 .
Question:3 Find the principal values of the following:
Answer:
Let us assume that , then we have;
, or
.
And we know the range of principal values is
Therefore the principal value of is .
Question:4 Find the principal values of the following:
Answer:
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 .
Question:5 Find the principal values of the following:
Answer:
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 :
Answer:
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 :
Answer:
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., .
Question:8 Find the principal values of the following:
Answer:
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
Question:9 Find the principal values of the following:
Answer:
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:
Answer:
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:
Answer:
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:
Answer:
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
Answer:
Given if then,
As we know that the can take values between
Therefore, .
Hence answer choice (B) is correct.
Question:14 is equal to
Answer:
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:
Answer:
Given to prove:
where, .
Take or
Take R.H.S value
=
=
=
= = L.H.S
Question:2 Prove the following:
Answer:
Given to prove .
Take or ;
Then we have;
R.H.S.
=
=
=
= = L.H.S
Hence Proved.
Question:4 Prove the following:
Answer:
Given to prove
Then taking L.H.S.
We have
= R.H.S.
Hence proved.
Question:5 Write the following functions in the simplest form:
Answer:
We have
Take
is the simplified form.
Question:7 Write the following functions in the simplest form:
Answer:
Given that
We have in inside the root the term :
Put and ,
Then we have,
Hence the simplest form is
Question:8 Write the following functions in the simplest form:
Answer:
Given where
So,
Taking common from numerator and denominator.
We get:
= as,
= is the simplest form.
Question:9 Write the following functions in the simplest form:
Answer:
Given that
Take or
and putting it in the equation above;
is the simplest form.
Question:10 Write the following functions in the simplest form:
Answer:
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:
Answer:
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:
Answer:
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
Answer:
Taking the value or and or then we have,
= ,
=
Then,
Ans.
Question:14 If , then find the value of .
Answer:
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 .
Answer:
Using the identity ,
We can find the value of x;
So,
on applying,
=
= or ,
Hence, the possible values of x are .
Question:16 Find the values of each of the expressions in Exercises 16 to 18.
Answer:
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
Question:17 Find the values of each of the expressions in Exercises 16 to 18.
Answer:
As we know
If which is the principal value range of .
So, as in ;
Hence we can write as :
=
Where
and
Question:18 Find the values of each of the expressions in Exercises 16 to 18.
Answer:
Given that
we can take ,
then
or
We have similarly;
Therefore we can write
from
Question:19 is equal to
Answer:
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
Answer:
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
Answer:
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:
Answer:
If then , which is principal value of .
So, we have
Therefore we have,
.
Question:2 Find the value of the following:
Answer:
We have given ;
so, as we know
So, here we have .
Therefore we can write as:
.
Question:3 Prove that
Answer:
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
Answer
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
Answer:
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
Answer:
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
Answer:
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
Answer:
Applying the formlua:
on two parts.
we will have,
Hence it s equal to R.H.S
Proved.
Question:9 Prove that
Answer:
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
Answer:
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
Answer:
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
Answer:
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:
Answer:
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:
Answer:
Given equation is
:
L.H.S can be written as;
Using the formula
So, we have
Hence the value of .
Question:16 then is equal to
Answer:
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).
NCERT solutions for class 12 maths chapterwise
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 

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 

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
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!!!