Question:
Solve for x :
$\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$
Solution:
Given: $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$
We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
So, $\sin ^{-1} \mathrm{x}=\frac{\pi}{2}-\cos ^{-1} \mathrm{X}$
Substituting in the given equation,
$\frac{\pi}{2}-\cos ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$
Rearranging,
$2 \cos ^{-1} x=\frac{\pi}{2}-\frac{\pi}{6}$
$2 \cos ^{-1} x=\frac{\pi}{3}$
$\cos ^{-1} x=\frac{\pi}{6}$
$x=\frac{\sqrt{3}}{2}$
Therefore, $x=\frac{\sqrt{3}}{2}$ is the required value of $x$.
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