Answer
$-\dfrac{\sqrt 3}{ 3}$
Work Step by Step
The range of $\cos^{-1}{x}$ is $[0, \pi]$.
Here, $\cos^{-1}{(-\dfrac{\sqrt 3}{2})}=\dfrac{ 5\pi}{6}$, because $\cos{(\dfrac{\pi}{6})}=-\dfrac{\sqrt 3}{2}$ and $\dfrac{5\pi}{6}$ lies in the range of $\cos^{-1}{x}$ .
Therefore, we have:
$\tan [\cos^{-1}{(\dfrac{\sqrt 3}{2})}]=\tan ({\dfrac{ 5 \pi}{6}})=-\dfrac{\sqrt 3}{ 3}$