Inverse Trigonometric Functions — Class 12 Maths Solution

${\tan ^{ - 1}}\left( {\tan \frac{{7\pi }}{6}} \right) \ne \frac{{7\pi }}{6}$ as the range of principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

So, ${\tan ^{ = 1}}\left( {\tan \frac{{7\pi }}{6}} \right) = {\tan ^{ - 1}}\left\{ {\tan \left( {\pi + \frac{\pi }{6}} \right)} \right\} = {\tan ^{ - 1}}\left( {\tan \frac{\pi }{6}} \right) = \frac{\pi }{6}$

$\therefore$ ${\tan ^{ - 1}}\left( {\tan \frac{{7\pi }}{6}} \right) = \frac{\pi }{6}$