The equation sin^{-1}x - cos^{-1}x = cos^{-1} | vedclass

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

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(B) Given the equation: $\sin^{-1}x - \cos^{-1}x = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$.We know that $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.So,the equation becomes: $\sin^{-1}x - \cos^{-1}x = \frac{\pi}{6}$.We also know the identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$.Adding these two equations:$(\sin^{-1}x - \cos^{-1}x) + (\sin^{-1}x + \cos^{-1}x) = \frac{\pi}{6} + \frac{\pi}{2}$.$2\sin^{-1}x = \frac{\pi + 3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.$\sin^{-1}x = \frac{\pi}{3}$.$x = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.Since we found exactly one value for $x$,the equation has a unique solution.

The equation $\sin^{-1}x - \cos^{-1}x = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$ has

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