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

Now, let $x=\tan ^{-1}\left(\tan \left(\frac{7 \pi}{6}\right)\right)$

$\Rightarrow \tan x=\tan \left(\frac{7 \pi}{6}\right)$

Here range of principle value of $\tan$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$\Rightarrow x=\frac{7 \pi}{6} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Hence for all values of $x$ in range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the value of

$\tan ^{-1}\left(\tan \left(\frac{13 \pi}{6}\right)\right)$ is

$\Rightarrow \tan x=\tan \left(\pi+\frac{\pi}{6}\right)\left(\because \tan \left(\frac{7 \pi}{6}\right)=\tan \left(\pi+\frac{\pi}{6}\right)\right)$

$\Rightarrow \tan x=\tan \left(\frac{\pi}{6}\right)(\because \tan (\pi+\theta)=\tan \theta)$

$\Rightarrow x=\frac{\pi}{6}$