Answer
$\dfrac{-\sqrt 3}{3}$
Work Step by Step
The range of $\sin^{-1}{x}$ is $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$.
Here, $\sin^{-1}{(\dfrac{-1}{2})}=\dfrac{-\pi}{6}$, because $\sin{(\dfrac{-\pi}{6})}=\dfrac{-1}{2}$ and $\dfrac{-\pi}{6}$ lies in the range of $\sin^{-1}{x}$ .
Therefore, we have:
$\tan [\sin^{-1}{(\dfrac{-1}{2})}]=\tan ({\dfrac{-\pi}{6}})=\dfrac{-\sqrt 3}{3}$
