Answer
$-\displaystyle \frac{1}{\sqrt{3}}$
Work Step by Step
$y=\sin^{-1}x$ is the number in $[-\pi/2, \pi/2]$ for which $\sin y=x.$
Sine is an odd function, so $\displaystyle \sin(\frac{\pi}{6})=\frac{1}{2}$ (in Q I)
which leads to $\displaystyle \sin(-\frac{\pi}{6})=-\frac{1}{2}$,
$\displaystyle \tan(\sin^{-1}(-\frac{1}{2}))=\tan(-\frac{\pi}{6})=$
We note that tan is also an odd function; thus:
$=-\displaystyle \tan\frac{\pi}{6}$
$=-\displaystyle \frac{1}{\sqrt{3}}$
