The value of sin left(cos ^{-1}left(-frac{1}{ | vedclass

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(A) We know that $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$ and $\sin ^{-1}(x) + \cos ^{-1}(x) = \frac{\pi}{2}$.Let the given expression be $E = \sin \lef

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$1$

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(A) We know that $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$ and $\sin ^{-1}(x) + \cos ^{-1}(x) = \frac{\pi}{2}$.Let the given expression be $E = \sin \left(\cos ^{-1}\left(-\frac{1}{3}\right)-\sin ^{-1}\left(\frac{1}{3}\right)\right)$.Using the property $\cos ^{-1}\left(-\frac{1}{3}\right) = \pi - \cos ^{-1}\left(\frac{1}{3}\right)$,we get:$E = \sin \left(\pi - \cos ^{-1}\left(\frac{1}{3}\right) - \sin ^{-1}\left(\frac{1}{3}\right)\right)$.Now,group the inverse trigonometric terms:$E = \sin \left(\pi - \left(\cos ^{-1}\left(\frac{1}{3}\right) + \sin ^{-1}\left(\frac{1}{3}\right)\right)\right)$.Since $\cos ^{-1}(x) + \sin ^{-1}(x) = \frac{\pi}{2}$ for $x \in [-1, 1]$,we have:$E = \sin \left(\pi - \frac{\pi}{2}\right) = \sin \left(\frac{\pi}{2}\right)$.Thus,$E = 1$.

The value of $\sin \left(\cos ^{-1}\left(-\frac{1}{3}\right)-\sin ^{-1}\left(\frac{1}{3}\right)\right)$ is

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