If $\sin ^{-1} \mathrm{x}+\sin ^{-1} \mathrm{y}=\frac{2 \pi}{3}$ then $\left(\cos ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{y}\right)=?$ A. $\frac{\pi}{6}$ B. $\frac{\pi}{3}$ C. $\pi$ D. $\frac{2 \pi}{3}$

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If $\sin ^{-1} \mathrm{x}+\sin ^{-1} \mathrm{y}=\frac{2 \pi}{3}$ then $\left(\cos ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{y}\right)=?$
A. $\frac{\pi}{6}$
B. $\frac{\pi}{3}$
C. $\pi$
D. $\frac{2 \pi}{3}$

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

If $\sin ^{-1} \mathrm{x}+\sin ^{-1} \mathrm{y}=\frac{2 \pi}{3}$ then $\left(\cos ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{y}\right)=?$
A. $\frac{\pi}{6}$
B. $\frac{\pi}{3}$
C. $\pi$
D. $\frac{2 \pi}{3}$

Solution:

Option(B) is correct.
Given: $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$
To Find: The value of $\cos ^{-1} x+\cos ^{-1} y$
Since we know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$\Rightarrow \cos ^{-1} x=\frac{\pi}{2}=\sin ^{-1} x$
Similarly $\cos ^{-1} y=\frac{\pi}{2}-\sin ^{-1} y$
Now consider $\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}-\sin ^{-1} x+\frac{\pi}{2}-\sin ^{-1} y$
$\begin{array}{l} =\frac{2 \pi}{2}-\left[\sin ^{-1} x+\sin ^{-1} y\right] \\ =\pi-\frac{2 \pi}{3} \end{array}$
$=\frac{\pi}{3}$