Question

Find the exact value of tan^-1 (-root 3/3)answer in radians in terms of π

Answer 1

To find $\tan^{-1}\left(-\dfrac{ \sqrt{3} }{3}\right)$ start with range of the function $y=\tan^{-1}x$. This function has range $y\in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.

Since $-\dfrac{ \sqrt{3} }{3}\ \textless \ 0$, then the angle $\tan^{-1}\left(-\dfrac{ \sqrt{3} }{3}\right)$ lies in VIth quadrant. 

It is well known that $\tan \dfrac{\pi}{6}=\dfrac{ \sqrt{3} }{3}$, then $\tan^{-1}\left(-\dfrac{ \sqrt{3} }{3}\right)=-\dfrac{\pi}{6}$ (the angle $-\dfrac{\pi}{6}$ lies in the VIth quadrant).