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प्रश्न
If \[y = se c^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right), x > 0 . \text{ Find} \frac{dy}{dx}\] ?
उत्तर
\[\text{ Here, y }= se c^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right)\]
\[ \Rightarrow y = \cos^{- 1} \left( \frac{x - 1}{x + 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right) ..........\left[ \text{ Since,} se c^{- 1} \left( x \right) = \cos^{- 1} \left( \frac{1}{x} \right) \right]\]
\[ \Rightarrow y = \frac{\pi}{2} ..........\left[ \text{ Since}, \cos^{- 1} \alpha + \sin^{- 1} \alpha = \frac{\pi}{2} \right]\]
Differentiate it with respect to x,
\[\therefore \frac{d y}{d x} = 0\]
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