Inverse Trigonometric Function - Practice Questions & MCQ

The trigonometric are many-one functions in their actual domain. For the inverse of trigonometric functions to be defined, the actual domain of trigonometric function must be restricted to make it one-one and onto function 

$y=\sin ^{-1}(x)$

The function y = sin(x) is many-one so it is not invertible. Now consider the small portion of the function 

$\mathrm{y}=\sin \mathrm{x}, \mathrm{x} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\mathrm{co}-$ domain $=[-1,1]$

For this domain and co-domain this function is one-one and onto, so it is invertible and its inverse is $y=\sin ^{-1}(x)$

Its Domain is $[-1,1]$ and Range is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$y=\cos ^{-1}(x)$

Domain is $[-1,1]$ and Range is $[0, \pi]$

$y=\tan ^{-1}(x)$

Domain is $\mathbb{R}$ and Range is $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

$y=\cot ^{-1}(x)$

Domain is $\mathbb{R}$ and Range is $(0, \pi)$

$y=\sec ^{-1}(x)$

Domain is $\mathbb{R}-(-1,1)$ and Range is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$

$y=\operatorname{cosec}^{-1}(x)$

Domain is $\mathbb{R}-(-1,1)$ and Range is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$