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If \(\sec^{-1} \left [ \frac {x+1}{x-1} \right ] + \sin^{-1} \left [ \frac {x-1}{x+1} \right ]\) then dy/dx is equal to
Concept:Inverse trigonometric identities\(\sin^{-1} (\frac 1 x) = \csc^{-1} (x) \)csc-1 (x) + sec-1 (x) = \(\frac \pi 2\)Calculation:Given:y = \(\sec^{-1} \left [ \frac {x+1}{x-1} \right ] + \sin^{-1} \left [ \frac {x-1}{x+1} \right ] = \sec^{-1} \left [ \frac {x+1}{x-1} \right ] + \sin^{-1} \left [ \frac {1}{\frac {x+1}{x-1}} \right ]\)= \(\sec^{-1} \left [ \frac {x+1}{x-1} \right ] + \csc^{-1} \left [ \frac {x+1}{x-1} \right ]\)y = \(\frac {\pi }{2}\)\(\frac {dy}{dx} = \frac {d}{dx} \frac {\pi }{2} = 0\)
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