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