Evaluate each of the following:

Solution:

We know that

$\tan ^{1}(\tan \theta)=\theta, \quad-\frac{\pi}{2}<\theta<\frac{\pi}{2}$

(i) We have

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

(ii) We have

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

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

$=-\frac{\pi}{7}$

(iii) We have

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

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

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

(iv) We have

$\tan ^{-1}\left(\tan \frac{9 \pi}{4}\right)=\tan ^{-1}\left[\tan \left(2 \pi+\frac{\pi}{4}\right)\right]$

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

$=\frac{\pi}{4}$

(v) We have

$\tan ^{-1}(\tan 1)=1$

(vi) We have

$\tan ^{-1}(\tan 2)=\tan ^{-1}[\tan (-\pi+2)]$

$=2-\pi$

(vii) We have

$\tan ^{-1}(\tan 4)=\tan ^{-1}[\tan (-\pi+4)]$

$=4-\pi$

(viii) We have

$\tan ^{-1}(\tan 12)=\tan ^{-1}[\tan (-4 \pi+12)]$

$=12-4 \pi$