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Question:

If $\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{3}$, then $\sin ^{-1} x+\sin ^{-1} y=$ ____________________.

Solution:

We know

$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$, for all $x \in \mathrm{R}$                           ...(1)

Also, $\sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}$, for all $y \in \mathrm{R}$                   ...(2)

Adding (1) and (2), we get

$\sin ^{-1} x+\cos ^{-1} x+\sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}+\frac{\pi}{2}$

$\Rightarrow \sin ^{-1} x+\sin ^{-1} y+\cos ^{-1} x+\cos ^{-1} y=\pi$

$\Rightarrow \sin ^{-1} x+\sin ^{-1} y+\frac{\pi}{3}=\pi \quad$ (Given)

$\Rightarrow \sin ^{-1} x+\sin ^{-1} y=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$

If $\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{3}$, then $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$

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