Evaluate each of the following:

Solution:

We know that

$\cot ^{-1}(\cot \theta)=\theta, \quad(0, \pi)$

(i) We have

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

(ii) We have

$\cot ^{-1}\left(\cot \frac{4 \pi}{3}\right)=\cot ^{-1}\left[\cot \left(\pi+\frac{\pi}{3}\right)\right]$

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

$=\frac{\pi}{3}$

(iii) We have

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

$=\cot ^{-1}\left(\cot \frac{\pi}{4}\right)$

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

(iv) We have

$\cot ^{-1}\left(\cot \frac{19 \pi}{6}\right)=\cot ^{-1}\left[\cot \left(\pi+\frac{\pi}{6}\right)\right]$

$=\cot ^{-1}\left(\cot \frac{\pi}{6}\right)$

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

(v) We have

$\cot ^{-1}\left[\cot \left(-\frac{8 \pi}{3}\right)\right]=\cot ^{-1}\left[-\cot \left(\frac{8 \pi}{3}\right)\right]$

$=\cot ^{-1}\left[-\cot \left(3 \pi-\frac{\pi}{3}\right)\right]$

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

$=\frac{\pi}{3}$

(vi) We have

$\cot ^{-1}\left(\cot \frac{21 \pi}{4}\right)=\cot ^{-1}\left[\cot \left(5 \pi+\frac{\pi}{4}\right)\right]$

$=\cot ^{-1}\left(\cot \frac{\pi}{4}\right)$

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