Prove that ^ -1 ( 1 + + 1 - 1 + - 1 - )= x 2 ; x (0, 4 ). - Vidyadip

Question

Prove that $\cot^{-1}\bigg(\frac{\sqrt{1 +\sin\text{x}} + \sqrt{1 -\sin\text{x}}}{\sqrt{1 + \sin\text{x}} - \sqrt{1 - \sin\text{x}}} \bigg)= \frac{\text{x}}{2}; \text{x}\in\bigg(0,\frac{\pi}{4}\bigg).$

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Answer

$\cot^{-1}\left\{\frac{\sqrt{1+ \sin\text{x}} + \sqrt{1 -\sin\text{x}}}{\sqrt{1 + \sin\text{x}} - \sqrt{1 -\sin\text{x}}}\right\}$
$ = \cot^{-1}\left\{\frac{\sqrt{\bigg(\cos\frac{\text{x}}{2} + \sin\frac{\text{x}}{2}\bigg)^{2}} + \sqrt{\bigg(\cos\frac{\text{x}}{2} - \sin\frac{\text{x}}{2}\bigg)}^{2}}{\sqrt{\bigg(\cos\frac{\text{x}}{2} + \sin\frac{\text{x}}{2}\bigg)^{2}} - \sqrt{\bigg(\cos\frac{\text{x}}{2} - \sin\frac{\text{x}}{2}\bigg)}^{2}}\right\}$
$ = \cot^{-1}\left\{\frac{2\cos\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}}\right\} = \cot^{-1}\bigg(\cot\frac{\text{x}}{2}\bigg) = \frac{\text{x}}{2}.$

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