# Inverse Trigonometric Functions Class 12 Notes Maths Chapter 2 - CBSE

## Inverse Function

Let f be a one-to-one correspondence from the set A to the set B. The inverse function of f is the function that assigns to an element b belonging to B the unique element a in A such that f(a)=b. The inverse function of f is denoted by f -1. Hence, f -1(b)=a when f(a)=b. Inverse trigonometric functions are simply defined as the inverse functions of basic trigonometric functions.

## Domains And Ranges Of Inverse Trigonometric Functions

The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table:

 Functions Domain Range (Principal Value Branches) y = sin -1 x [-1,1] [—ℼ/2 , ℼ/2] y = cos -1 x [-1,1] [0, ℼ] y = cosec -1 x R-(-1,1) [—ℼ/2 , ℼ/2] - {0} y = sec -1 x R-(-1,1) [0, ℼ] - {ℼ/2} y = tan -1 x R (—ℼ/2 , ℼ/2) y = cot -1 x R (0, ℼ)

Domain of inverse trigonometry function = Range of trigonometry function.

## Graphs Of Inverse Trigonometric Functions

(i) Inverse sine function

In,[—ℼ/2 , ℼ/2], sinx is bijective hence its inverse is y = sin-1 x, x[-1,1] and y [—ℼ/2 , ℼ/2]. (ii) Inverse cosine function

In, [0,ℼ] cosine function is bijective and hence its inverse is y = cos-1 x, x [-1,1] and y [0,]. (iii) Inverse tangent function

Tangent function is one-one and onto from (-ℼ/2 ,ℼ/2) to (-∞,∞) and so in this region the tangent function is invertible i.e., y = tan-1 x, x [-∞,∞] and y (-ℼ/2 ,ℼ/2) (iv) Inverse cotangent function

Cotangent function is one-one and onto from (0,ℼ) to (-∞,∞) hence cot x is invertible in this region i.e., y = cot-1 x; x [-∞,∞] and y (0,ℼ) (v) Inverse secant function

A function f : = [0, -ℼ/2) U (-ℼ/2, ℼ] → (-∞,-1] U [1,∞) defined by f(x) = sec x is one-one and onto and hence it is invertible in this region, hence y = sec-1 x; x(-∞,-1] U [1,∞) and y [0, ℼ/2) U (ℼ/2, ℼ] (vi) Inverse cosecant function

A function f : = [- ℼ/2, 0) U (0, ℼ/2] → (-∞,-1] U [1,∞) defined by
f(x) = cosec x is one-one and onto and hence it is invertible in this
region, hence y = cosec-1 x; x(-∞,-1] U [1,∞) and y [- ℼ/2, 0) U (0, ℼ/2] 