If y = sin^{-1}left(frac{x^2 - 1}{x^2 + 1}rig | vedclass

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(A) We know that $\sec^{-1}(u) = \cos^{-1}(\frac{1}{u})$ for $|u| \geq 1$.Given $y = \sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) + \sec^{-1}\left(\f

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(A) We know that $\sec^{-1}(u) = \cos^{-1}(\frac{1}{u})$ for $|u| \geq 1$.Given $y = \sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}ight) + \sec^{-1}\left(\frac{x^2 + 1}{x^2 - 1}ight)$.Using the identity $\sec^{-1}\left(\frac{x^2 + 1}{x^2 - 1}ight) = \cos^{-1}\left(\frac{x^2 - 1}{x^2 + 1}ight)$.Substituting this into the expression for $y$,we get $y = \sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}ight) + \cos^{-1}\left(\frac{x^2 - 1}{x^2 + 1}ight)$.Since $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$ for $|	heta| \leq 1$,and here $\left|\frac{x^2 - 1}{x^2 + 1}ight| Since $y$ is a constant,the derivative $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) = 0$.

If $y = \sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) + \sec^{-1}\left(\frac{x^2 + 1}{x^2 - 1}\right)$,$|x| > 1$,then $\frac{dy}{dx}$ is equal to:

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