Question

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

Options

• − 1/4

• − 1/2

• 1/4

• 1/2

Answer 1

1/2

\[\text{ Let y } = \tan^{- 1} \left\{ \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right\}\]
\[ \Rightarrow y = \tan^{- 1} \left\{ \sqrt{\frac{1 - \cos\left( \frac{\pi}{2} + x \right)}{1 + \cos\left( \frac{\pi}{2} + x \right)}} \right\}\]
\[ \Rightarrow y = \tan^{- 1} \left\{ \sqrt{\frac{2 \sin^2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{2 \cos^2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}} \right\} \]
\[ \Rightarrow y = \tan^{- 1} \left\{ \tan\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\} = \frac{\pi}{4} + \frac{x}{2}\]
\[ \therefore \frac{dy}{dx} = \frac{1}{2}\]