Inverse Trigonometric Functions — Class 12 Maths Solution

${\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right) \ne \frac{{3\pi }}{4}$ as the principal value branch of ${\tan ^{ - 1}}\;is\;\left( { - \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

So, ${\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}\left( {\tan \left( {\pi - \frac{\pi }{4}} \right)} \right)$

$= {\tan ^{ - 1}}\left[ { - \tan \left( {\frac{\pi }{4}} \right)} \right]$

$= {\tan ^{ - 1}}\left( {\tan \left( { - \frac{\pi }{4}} \right)} \right)$

$= - \frac{\pi }{4} \in \left( { - \frac{\pi }{2},\;\,\frac{\pi }{2}} \right)$

Hence, ${\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right) = - \frac{\pi }{4}.$