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Question:

If $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$, then $x=$_________________________.

Solution:

Given: $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$             …(1)

We know

$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$                             …(2)

Adding $(1)$ and $(2)$, we get

$2 \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}$

$\Rightarrow 2 \sin ^{-1} x=\frac{4 \pi}{6}$

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

$\Rightarrow x=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$

If $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$, then $x=\frac{\sqrt{3}}{2}$

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