Derivative of cot - 1 [ 1 + sin x + 1 - sin x 1 + sin x - 1 - sin x ] , 0 x 2 is:

Question

Derivative of $\cot^{- 1} [\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}], 0 < x < \frac{π}{2}$ is:

Smart Achievers · Accepted Answer

(1)

Ans. $\frac{1}{2}$

Explanation:

$Let y = \cot^{- 1} (\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}})$

$y = \cot^{- 1} [\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}]$

$\Rightarrow$ $y = \cot^{- 1} (\frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}}) = \cot^{- 1} (\cot \frac{x}{2}) = \frac{x}{2}$

$∴$ $\frac{d y}{d x} = \frac{1}{2}$

Solution

Q.
Derivative of $\cot^{- 1} [\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}], 0 < x < \frac{π}{2}$ is:

1 $\frac{1}{2}$

2 $1$

3 $2$

4 None of these

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