To find [tex]\tan^{-1}\left(-\dfrac{ \sqrt{3} }{3}\right)[/tex] start with range of the function [tex]y=\tan^{-1}x[/tex]. This function has range [tex]y\in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right][/tex].
Since [tex]-\dfrac{ \sqrt{3} }{3}\ \textless \ 0[/tex], then the angle [tex]\tan^{-1}\left(-\dfrac{ \sqrt{3} }{3}\right)[/tex] lies in VIth quadrant.
It is well known that [tex]\tan \dfrac{\pi}{6}=\dfrac{ \sqrt{3} }{3}[/tex], then [tex]\tan^{-1}\left(-\dfrac{ \sqrt{3} }{3}\right)=-\dfrac{\pi}{6}[/tex] (the angle [tex]-\dfrac{\pi}{6}[/tex] lies in the VIth quadrant).
