Answer
$-\dfrac{\sqrt3}{3}$
Work Step by Step
Note that $\cos^{-1}{\left(-\dfrac{\sqrt3}{2}\right)}=\dfrac{5\pi}{6}$, because $\cos{\left(\dfrac{5\pi}{6}\right)}=-\dfrac{\sqrt3}{2}$ and $d\frac{5\pi}{6}$ is in the range of $\cos^{-1}{x}$, which is $\left[0,\pi\right].$
Thus
$\tan{\left(\cos^{-1}{\left(\dfrac{\sqrt3}{2}\right)}\right)}=\tan{\dfrac{5\pi}{6}}=-\dfrac{\sqrt3}{3}$