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;

$\dpi{80} \sin z = (\frac{-1}{2}) = -\sin(\frac{\pi}{6})= \sin(\frac{-\pi}{6})$.

As we know that the range of principal values of $\sin^{-1}$ is from $\dpi{100} (\frac{-\pi}{2}, \frac{\pi}{2})$.

Hence only one possible value is $\dpi{80} \frac{-\pi}{6}$ .

Therefore, the principal value of $\dpi{80} \sin^{-1}\left(-\frac{1}{2} \right )$ is $\dpi{80} \frac{-\pi}{6}$.

Que.2 Find the principal values of:

$\dpi{80} \cos^{-1}\left(\frac{\sqrt3}{2} \right )$

Solution-

So, let us assume that $\dpi{80} \cos^{-1}\left(\frac{\sqrt3}{2} \right ) = x$

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 $\dpi{80} cos\ x = (\frac{\sqrt{3}}{2})$     when $\dpi{80} x =\frac{\pi}{6}$ .

Therefore, the principal value for $\dpi{80} \cos^{-1}\left(\frac{\sqrt3}{2} \right )$is   $\dpi{100} \frac{\pi}{6}$.

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 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