Mark the tick against the correct answer in the following:

To Find: The value of $\sec ^{-1}\left(\sec \left(\frac{8 \pi}{5}\right)\right)$

Now, let $x=\sec ^{-1}\left(\sec \left(\frac{8 \pi}{5}\right)\right)$

$\Rightarrow \sec x=\sec \left(\frac{8 \pi}{5}\right)$

Here range of principle value of sec is $[0, \pi]$

$\Rightarrow x=\frac{8 \pi}{5} \notin[0, \pi]$

Hence for all values of $x$ in range $[0, \pi]$, the value of

$\sec ^{-1}\left(\sec \left(\frac{8 \pi}{5}\right)\right)$ is

$\Rightarrow \sec x=\sec \left(2 \pi-\frac{2 \pi}{5}\right)\left(\because \sec \left(\frac{8 \pi}{5}\right)=\sec \left(2 \pi-\frac{2 \pi}{5}\right)\right)$

$\Rightarrow \sec x=\sec \left(\frac{2 \pi}{5}\right)(\because \sec (2 \pi-\theta)=\sec \theta)$

$\Rightarrow x=\frac{2 \pi}{5}$