# Mark the tick against the correct answer in the following

Question:

Mark the tick against the correct answer in the following

$\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=?$

A. $\frac{4 \pi}{3}$

B. $\frac{\pi}{2}$

C. $\frac{5 \pi}{3}$

D. $\pi$

Solution:

To Find: The value of $\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)+\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$

Here,consider $\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)$ (\because the principle value of $\cos$ lies in the range $[0, \pi]$ and since $\frac{2 \pi}{3} \in[0, \pi]$ )

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

Now, consider $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$

Since here the principle value of sine lies in range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and since $\frac{2 \pi}{3} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

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

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

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

Therefore,

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

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

$=\pi$