Domain and Range of Trigonometric Functions - Practice Questions & MCQ

Domain and range of Inverse Trigonometric Function (Part 1)

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

The function is many one so it is not invertible. Now consider the small portion of the function

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

Which is strictly increasing, Hence, one-one and inverse is $y=\sin ^{-1}(x)$

$\mathrm{Domain\;is\;[-1,1]\;\;and\;\;Range\;\;is\;\;\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]}$

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

$\mathrm{Domain\;is\;[-1,1]\;\;and\;\;Range\;\;is\;\;\left [ 0,\pi\right ]}$

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

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

Domain and range of Inverse Trigonometric Function (Part 2)

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

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

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

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

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

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