The value of cot−1⁡1−sin⁡x+1+sin⁡x1−sin⁡x−1+sin⁡x, is 0<x<π2

The value of cot−1⁡1−sin⁡x+1+sin⁡x1−sin⁡x−1+sin⁡x, is 0

A
$π - \frac{x}{2}$
B
$2 π - x$
C
$\frac{x}{2}$
D
$2 π - \frac{x}{2}$

Solution:

We have,

$1 \pm \sin x = {(\cos \frac{x}{2} \pm \sin \frac{x}{2})}^{2}$

$\begin{array}{l} \Rightarrow \sqrt{1 \pm \sin x} = \sqrt{{(\cos \frac{x}{2} \pm \sin \frac{x}{2})}^{2}} = c os \frac{x}{2} \pm \sin \frac{x}{2} \\ \Rightarrow \sqrt{1 \pm \sin x} = \cos \frac{x}{2} \pm \sin \frac{x}{2} \end{array}$

$∴ \cot^{- 1} \{\frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}\}$

$\begin{array}{l} = \cot^{- 1} \{\frac{(\cos \frac{x}{2} - \sin \frac{x}{2}) + (\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2}) - (\cos \frac{x}{2} + \sin \frac{x}{2})}\} \\ = \cot^{- 1} \{\cot (π - \frac{x}{2})\} = π - \frac{x}{2} \end{array}$

The value of $\cot^{- 1} \{\frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}\},$ is $(0 < x < \frac{π}{2})$

Solution:

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