#### Answer

$\sin^{-1}\sqrt{3}\ \ $ does not exist

#### Work Step by Step

$y=\sin^{-1}x =\arcsin x$
Domain: $[-1, 1] $
Range: $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$
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$y$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$
such that $\sin y=\sqrt{3}.$
Such a y can not exist, as
$-1 \leq \sin y \leq 1$
($\sqrt{3}$ is not in the range of $\sin x,$
$\sqrt{3}$ is not in the domain of $\sin^{-1}x$)
so
$\sin^{-1}\sqrt{3}$ does not exist