If sin ^{-1} x - cos ^{-1} x = frac{pi}{6},th | vedclass

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(B) Given equation: $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$.We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$.Adding the two equations:

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$\frac{\sqrt{3}}{2}$

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(B) Given equation: $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$.We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$.Adding the two equations:$(\sin ^{-1} x - \cos ^{-1} x) + (\sin ^{-1} x + \cos ^{-1} x) = \frac{\pi}{6} + \frac{\pi}{2}$$2 \sin ^{-1} x = \frac{\pi + 3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$$\sin ^{-1} x = \frac{\pi}{3}$$x = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

If $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$,then $x$ is equal to

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