Question:
$\sin ^{-1} x=\frac{\pi}{6}+\cos ^{-1} x$
Solution:
$\sin ^{-1} x=\frac{\pi}{6}+\cos ^{-1} x$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}-\sin ^{-1} x$ $\left[\because \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x\right]$
$\Rightarrow 2 \sin ^{-1} x=\frac{2 \pi}{3}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{3}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{3}$
$\Rightarrow \mathrm{x}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$
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