#### Answer

$\displaystyle \csc^{-1}\frac{\sqrt{2}}{2}$
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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$\displaystyle \frac{\sqrt{2}}{2}\approx 0.707\not\in(-\infty, -1]\cup[1, \infty)$.
So,
$\displaystyle \frac{\sqrt{2}}{2}$ 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 $\displaystyle \csc y=\frac{1}{\sin y}=\frac{\sqrt{2}}{2},$
as this would mean that $\displaystyle \sin y=\frac{2}{\sqrt{2}} > 1) $
$\displaystyle \csc^{-1}\frac{\sqrt{2}}{2}$ does not exist