Answer
$-\frac{\pi}{6}$
Work Step by Step
We need to find the angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.
$\frac{\sqrt 3}{3}=\frac{1}{\sqrt 3}=\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}=\frac{\sin\frac{\pi}{6}}{\cos\frac{\pi}{6}}=\tan\frac{\pi}{6}$
As $\tan(-x)=-\tan x$, $\tan(-\frac{\pi}{6})=-\frac{\sqrt 3}{3}$
The angle is $-\frac{\pi}{6}$.
$\tan^{-1} (-\frac{\sqrt 3}{3})=-\frac{\pi}{6}$