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If $\sin^{-1}x=y$ , then the range of $y$ is ,
Option: 1
$-\frac{\pi}{2}<y<\frac{\pi}{2}$

Option: 2
$-\frac{\pi}{2}\leq y\leq \frac{\pi}{2}$

Option: 3
$0\leq y\leq \frac{\pi}{2}$

Option: 4
$-\frac{\pi}{2}\leq y\leq 0$

Answers (1)

best_answer

Given that,

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

It is known that the range of $\sin$ is,

$\left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]$

Therefore,

$y\in \left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]$

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