Solution:

Option(C) is correct.
To Find: The value of $\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
Let, $x=\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
$\Rightarrow x=-\sin ^{-1}\left(\frac{1}{2}\right)+2\left[\pi-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right]\left(\because \sin ^{-1}(-\theta)=-\sin ^{-1}(\theta) \text { and } \cos ^{-1}(-\theta)=\pi-\cos ^{-1}(\theta)\right)$
$\begin{array}{l}
\Rightarrow x=-\left(\frac{\pi}{6}\right)+2\left[\pi-\frac{\pi}{6}\right] \\
\Rightarrow x=-\left(\frac{\pi}{6}\right)+2\left[\frac{5 \pi}{6}\right] \\
\Rightarrow x=-\frac{\pi}{6}+\frac{5 \pi}{3} \\
\Rightarrow x=\frac{3 \pi}{2}
\end{array}$