Question

Prove that $tan^{-1} (\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx} )-\sqrt{1-sinx}} \ ) = \frac{x}{2}$

Answer 1

${tan}^{ - 1} \frac{ \sqrt{1 + sinx} + \sqrt{1 - sinx} }{ \sqrt{1 + sinx} - \sqrt{1 - sinx} } \\ = {tan}^{ - 1} \frac{ \sqrt{( {sin \frac{x}{2} + cos \frac{x}{2} })^{2} } + \sqrt{( {sin \frac{x}{2} - cos \frac{x}{2} })^{2} } }{ \sqrt{ ({sin \frac{x}{2} + cos \frac{x}{2} })^{2} } - \sqrt{( {sin \frac{x}{2} - cos \frac{x}{2} })^{2} } } \\ = {tan}^{ - 1} \frac{sin \frac{x}{2} + cos \frac{x}{2} + sin \frac{x}{2} - cos \frac{x}{2} }{sin \frac{x}{2} + cos \frac{x}{2} - sin \frac{x}{2} + cos \frac{x}{2} } \\ = {tan}^{ - 1} \frac{2sin \frac{x}{2} }{2cos \frac{x}{2} } \\ = {tan}^{ - 1} tan \frac{x}{2} \\ = \frac{x}{2}$