Question: Find the value of $\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)$
Solution:
$\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)=\tan ^{-1}\left[\tan \left(\pi+\frac{\pi}{8}\right)\right]$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{8}\right)\right]$
$=\frac{\pi}{8}$
