Answer
$ \frac{\sqrt3}{2}$
Work Step by Step
RECALL:
(1) $\theta = \\csc^{-1}{(x)} \longrightarrow \csc{\theta}=x$
(2) $y=\csc^{-1}{x}$ is defined only in $[-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$, which means the angle is either in Quadrant I or IV.
(3) $\csc{\theta} = \frac{r}{y}$
(4) $\cos{\theta} = \frac{x}{r}$, where $r^2=x^2+y^2$
Let $\theta = \csc^{-1}{(-2)}$
This means that $\csc{\theta} = -2$. Since cosecant is negative, then the angle is in Quadrant IV.
With $\csc{\theta} = -2=\frac{2}{-1}=\frac{r}{y}$, we can set $y=-1$ and $r=2$.
Solve for $x$ using the formula $r^2=x^2+y^2$ to obtain:
$r^2=x^2+y^2
\\2^2=x^2+(-1)^2
\\4=x^2+1
\\4-1=x^2
\\3=x^2
\\\pm\sqrt3=x$
Since the angle is in Quadrant IV, then $x$ is positive so $x=\sqrt3$.
Thus,
$\cos{(\csc^{-1}{(-2)})}
\\=\cos{\theta}
\\= \frac{x}{r}
\\= \frac{\sqrt3}{2}$
