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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