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प्रश्न
If \[f\left( x \right) = \tan^{- 1} \sqrt{\frac{1 + \sin x}{1 - \sin x}}, 0 \leq x \leq \pi/2, \text{ then } f' \left( \pi/6 \right) \text{ is }\] _________ .
विकल्प
− 1/4
− 1/2
1/4
1/2
उत्तर
1/2
\[\text{ Let y } = \tan^{- 1} \left\{ \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right\}\]
\[ \Rightarrow y = \tan^{- 1} \left\{ \sqrt{\frac{1 - \cos\left( \frac{\pi}{2} + x \right)}{1 + \cos\left( \frac{\pi}{2} + x \right)}} \right\}\]
\[ \Rightarrow y = \tan^{- 1} \left\{ \sqrt{\frac{2 \sin^2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{2 \cos^2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}} \right\} \]
\[ \Rightarrow y = \tan^{- 1} \left\{ \tan\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\} = \frac{\pi}{4} + \frac{x}{2}\]
\[ \therefore \frac{dy}{dx} = \frac{1}{2}\]
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