Mark the tick against the correct answer in the following:
If $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$ then $\left(\cos ^{-1} x+\cos ^{-1} y\right)=?$
A. $\frac{\pi}{6}$
B. $\frac{\pi}{3}$
C. $\pi$
D. $\frac{2 \pi}{3}$
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 $\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}-\sin ^{-1} x+\frac{\pi}{2}-\sin ^{-1} y$
$=\frac{2 \pi}{2}-\left[\sin ^{-1} x+\sin ^{-1} y\right]$
$=\pi-\frac{2 \pi}{3}$
$=\frac{\pi}{3}$
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