Answer
$\displaystyle \csc^{-1}(-\frac{1}{2})$ does not exist
Work Step by Step
$y=\csc^{-1}x$
Domain: $(-\infty, -1]\cup[1, \infty)$
Range: $[-\displaystyle \frac{\pi}{2},0)\cup(0,\frac{\pi}{2}]$
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$-\displaystyle \frac{1}{2}\not\in(-\infty, -1]\cup[1, \infty)$.
So,
$-\displaystyle \frac{1}{2}$ is not in the domain of $\csc^{-1}x$,
meaning that it is not in the range of $\csc x.$
(There is no y from $[-\displaystyle \frac{\pi}{2},0)\cup(0,\frac{\pi}{2}]$
such that $\displaystyle \csc y=\frac{1}{\sin y}=-\frac{1}{2},$
as this would mean that $\sin y=-2 < -1) $
$\displaystyle \csc^{-1}(-\frac{1}{2})$ does not exist
