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Question
If \[y = \sec^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right)\] then write the value of \[\frac{dy}{dx} \] ?
Solution
\[\text{ We have, y } = \sec^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right) \]
\[ \Rightarrow y = \cos^{- 1} \left( \frac{x - 1}{x + 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right) ........\left[ \because \sec^{- 1} x = \cos^{- 1} \left( \frac{1}{x} \right) \right]\]
\[ \Rightarrow y = \frac{\pi}{2} ..........\left[ \because \sin^{- 1} x + \cos^{- 1} x = \frac{\pi}{2} \right]\]
\[\Rightarrow \frac{dy}{dx} = 0\]
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