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. $\pi$

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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. $\pi$

CBSE | Class 12 | Inverse Trigonometric Functions | Maths | RS Aggarwal

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:

Option(D) is correct.
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 $\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)\left(\because\right.$ the principle value of $\cos$ lies in the range $[0, \pi]$ and since $\left.\frac{2 \pi}{3} \in[0, \pi]\right)$ $\Rightarrow \cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)=\frac{2 \pi}{3}$
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 $\frac{2 \pi}{3} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\begin{array}{l} \Rightarrow \sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{3}\right)\right) \\ =\sin ^{-1}\left(\sin \left(\frac{\pi}{3}\right)\right) \\ =\frac{\pi}{3} \end{array}$
Therefore,
$\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)+\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)=\frac{2 \pi}{3}+\frac{\pi}{3}$
$=\frac{3 \pi}{3}$
$=\pi$