Inverse Trigonometric Functions — Class 12 Maths Solution

We have, ${\tan ^{ - 1}}\left( {\tan \frac{{2\pi }}{3}} \right) = {\tan ^{ - 1}}\tan \left( {\pi - \frac{\pi }{3}} \right)$

$= {\tan ^{ - 1}}\left( { - \tan \frac{\pi }{3}} \right)$

$= - {\tan ^{ - 1}}\tan \frac{\pi }{3} = - \frac{\pi }{3}$

Note
Remember that, ${\tan ^{ - 1}}\left( {\tan \frac{{2\pi }}{3}} \right) \ne \frac{{2\pi }}{3}$

Since, ${\tan ^{ - 1}}(\tan x) = x$, if $x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$

and $\frac{{2\pi }}{3} \notin \left( {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right)$