Question

Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .

Answer 1

\[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\]

\[ = \cot^{- 1} \left\{ \frac{\sqrt{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^2} + \sqrt{\left( \cos\frac{x}{2} - \sin\frac{x}{2} \right)^2}}{\sqrt{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^2} - \sqrt{\left( \cos\frac{x}{2} - \sin\frac{x}{2} \right)^2}} \right\}\]

\[\left[ \because \left( \cos\frac{x}{2} \pm \sin\frac{x}{2} \right)^2 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} \pm 2\sin\frac{x}{2}\cos\frac{x}{2} = 1 \pm \sin x \right]\]

\[ = \cot^{- 1} \left\{ \frac{\left| \cos\frac{x}{2} + \sin\frac{x}{2} \right| + \left| \cos\frac{x}{2} - \sin\frac{x}{2} \right|}{\left| \cos\frac{x}{2} + \sin\frac{x}{2} \right| - \left| \cos\frac{x}{2} - \sin\frac{x}{2} \right|} \right\} \]

\[ = \cot^{- 1} \left\{ \frac{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right) + \left( \cos\frac{x}{2} - \sin\frac{x}{2} \right)}{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right) - \left( \cos\frac{x}{2} - \sin\frac{x}{2} \right)} \right\} \left[ \because 0 < \frac{x}{2} < \frac{\pi}{4} \therefore \cos\frac{x}{2} > \sin\frac{x}{2} \right]\]

\[ = \cot^{- 1} \left( \cot\frac{x}{2} \right)\]

\[ = \frac{x}{2}\]