Find the value of


Find the value of $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$


We know that $\tan ^{-1}(\tan x)=x$ if $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, which is the principal value branch of $\tan ^{-1} x$.

Here, $\frac{7 \pi}{6} \notin\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Now, $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$ can be written as:

$\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\tan ^{-1}\left[\tan \left(2 \pi-\frac{5 \pi}{6}\right)\right]$    $[\tan (2 \pi-x)=-\tan x]$

$=\tan ^{-1}\left[-\tan \left(\frac{5 \pi}{6}\right)\right]=\tan ^{-1}\left[\tan \left(-\frac{5 \pi}{6}\right)\right]=\tan ^{-1}\left[\tan \left(\pi-\frac{5 \pi}{6}\right)\right]$

$=\tan ^{-1}\left[\tan \left(\frac{\pi}{6}\right)\right]$, where $\frac{\pi}{6} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

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

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