Answer
$\frac{\pi}{6}$
Work Step by Step
$\tan \frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})=\tan\frac{\pi}{6}$
($\tan(\pi+x)=\tan x$)
Therefore, $\tan^{-1}(\tan \frac{7\pi}{6})=\tan^{-1}(\tan \frac{\pi}{6})$
Within the restricted interval ($-\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$),
$\tan^{-1}(\tan \theta)=\theta$
$\implies \tan^{-1}(\tan \frac{7\pi}{6})=\tan^{-1}(\tan\frac{\pi}{6})=\frac{\pi}{6}$
