Answer
$-\dfrac{\pi}{6}$
Work Step by Step
Suppose $\theta = \tan^{-1} (-\dfrac{\sqrt 3}{3})$
This gives: $\tan \theta= -\dfrac{\sqrt 3}{3}$
Now, $\theta = \tan^{-1} (-\dfrac{\sqrt 3}{3})=-\dfrac{\pi}{6}$
Therefore, $\tan^{-1} (-\dfrac{\sqrt 3}{3})=-\dfrac{\pi}{6}$
