NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

 

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

In chapter 1 we already study that if a function(f) is one-one and onto then the inverse of that function exists which is denoted f^{-1}. In this chapter, we are going to study the inverse of trigonometric functions and properties of inverse trigonometric functions. It is a very important chapter for the board exam and also for the competitive exam. In this chapter, there are 2 exercises with 35 questions. The NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions are prepared and explained by the subject experts.

In mathematics Inverse Trigonometric Functions(ITF) are the inverse function of the trigonometric function. It is also termed as cyclometric functions, arcus functions or anti trigonometric functions. 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\sin^-^1x=\frac{1}{\sin x}  . But the inverse function (f^-^1) is not the same as  \frac{1}{f} for example \sin^-^1x\neq \frac{1}{\sin x}.

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

Some Examples are-

Que.1 Find the principal value of-

     \sin^-^1(\frac{-1}{2})

Solution-

Let us assume that \sin^-^1(\frac{-1}{2})=

Then taking inverse both sides we get;

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

As we know that the range of principal values of \sin^{-1} is from (\frac{-\pi}{2}, \frac{\pi}{2}).

Hence only one possible value is \frac{-\pi}{6} .

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

 

Que.2 Find the principal values of:

\cos^{-1}\left(\frac{\sqrt3}{2} \right )

Solution-

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

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}.

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

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

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.2 

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

NCERT Solutions for class 12- Maths

Chapter 1

Relations and Functions

Chapter 3

Matrices

Chapter 4

Determinants

Chapter 5

Continuity and Differentiability

Chapter 6

Application of Derivatives

Chapter 7

Integrals

Chapter 8

Application of Integrals

Chapter 9

Differential Equations

Chapter 10

Vector Algebra

Chapter 11

Three Dimensional Geometry

Chapter 12

Linear Programming

Chapter 13

Probability

NCERT Solutions for Class 12

 

Boost Your Preparation with these Knockout courses

 

Recently Asked Questions

 

Related Articles

Exams
Articles
Questions