Find the value
Question:

Find the value of $\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)$

Solution:

$\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)=\tan ^{-1}\left[\tan \left(\pi+\frac{\pi}{8}\right)\right]$

$=\tan ^{-1}\left[\tan \left(\frac{\pi}{8}\right)\right]$

$=\frac{\pi}{8}$

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