Answer
$\sin^{-1}\sqrt{3}\ \ \ $ is not defined.
Work Step by Step
Inverse Sine Function:
$y=\sin^{-1}x$ or $y=$ arcsin $x$ means that
$x=\sin y$, for $-\displaystyle \frac{\pi}{2} \leq y \leq \frac{\pi}{2}$.
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Since $\sqrt{3} > 1$, there is no y such that
$\sin y=\sqrt{3}$
$y=\sin^{-1}\sqrt{3}$ is not defined.
($\sin^{-1}$ is the inverse of $\sin,$ so its domain must be the range of sine.
The range of sine is $[-1,1].$
$\sqrt{3}\not\in [-1,1] )$
