Evaluate each of the following:

Solution:

We know that

$\sec ^{1}(\sec \theta)=\theta, \quad[0, \pi / 2) \cup(\pi / 2, \pi]$

(i) We have

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

(ii) We have

$\sec ^{-1}\left(\sec \frac{2 \pi}{3}\right)=\frac{2 \pi}{3}$

(iii) We have

$\sec ^{-1}\left(\sec \frac{5 \pi}{4}\right)=\sec ^{-1}\left[\sec \left(2 \pi-\frac{3 \pi}{4}\right)\right]$

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

$=\frac{3 \pi}{4}$

(iv)We have

$\sec ^{-1}\left(\sec \frac{7 \pi}{3}\right)=\sec ^{-1}\left[\sec \left(2 \pi+\frac{\pi}{3}\right)\right]$

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

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

(v)We have

$\sec ^{-1}\left(\sec \frac{9 \pi}{5}\right)=\sec ^{-1}\left[\sec \left(2 \pi-\frac{\pi}{5}\right)\right]$

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

$=\frac{\pi}{5}$

(vi) We have

$\sec ^{-1}\left\{\sec \left(-\frac{7 \pi}{3}\right)\right\}=\sec ^{-1}\left\{\sec \left(\frac{7 \pi}{3}\right)\right\}$

$=\sec ^{-1}\left[\sec \left(2 \pi+\frac{\pi}{3}\right)\right]$

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

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

(vii)We have

$\sec ^{-1}\left(\sec \frac{13 \pi}{4}\right)=\sec ^{-1}\left[\sec \left(4 \pi-\frac{3 \pi}{4}\right)\right]$

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

$=\frac{3 \pi}{4}$

(viii)We have

$\sec ^{-1}\left(\sec \frac{25 \pi}{6}\right)=\sec ^{-1}\left[\sec \left(4 \pi+\frac{\pi}{6}\right)\right]$

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

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