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Solution for If F ( X ) = Tan − 1 √ 1 + Sin X 1 − Sin X , 0 ≤ X ≤ π / 2 , Then F ′ ( π / 6 ) is (A) − 1/4 (B) − 1/2 - CBSE (Commerce) Class 12 - Mathematics

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Question

If \[f\left( x \right) = \tan^{- 1} \sqrt{\frac{1 + \sin x}{1 - \sin x}}, 0 \leq x \leq \pi/2, \text{ then } f' \left( \pi/6 \right) \text{ is }\]

(a) − 1/4
(b) − 1/2
(c) 1/4
(d) 1/2

Solution

(d) 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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Solution If F ( X ) = Tan − 1 √ 1 + Sin X 1 − Sin X , 0 ≤ X ≤ π / 2 , Then F ′ ( π / 6 ) is (A) − 1/4 (B) − 1/2 Concept: Simple Problems on Applications of Derivatives.
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