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To Find: The value of $\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$

Let, $x=\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$

$\Rightarrow x=-\sin ^{-1}\left(\frac{1}{2}\right)+2\left[\pi-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right]\left(\because \sin ^{-1}(-\theta)=-\sin ^{-1}(\theta)\right.$ and $\left.\cos ^{-1}(-\theta)=\pi-\cos ^{-1}(\theta)\right)$

$\Rightarrow x=-\left(\frac{\pi}{6}\right)+2\left[\pi-\frac{\pi}{6}\right]$

$\Rightarrow x=-\left(\frac{\pi}{6}\right)+2\left[\frac{5 \pi}{6}\right]$

$\Rightarrow x=-\frac{\pi}{6}+\frac{5 \pi}{3}$

$\Rightarrow x=\frac{3 \pi}{2}$

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