Question

Cot inverse{√ 1+ sinx+ √ 1- sinx}/{ √1+sinx-√1-sinx} =2/3

Answer 1

${cot}^{ - 1} ( \frac{ \sqrt{1 + sinx} + \sqrt{1 - sinx} }{ \sqrt{1 + sinx} - \sqrt{ 1 - sinx} } ) = \frac{2}{3} \\ \\ {cot}^{ - 1} ( \frac{ \sqrt{( {cos \frac{x}{2} + sin \frac{x}{2} })^{2} } + \sqrt{( {cos \frac{x}{2} - sin \frac{x}{2} })^{2} } }{ \sqrt{( {cos \frac{x}{2} + sin \frac{x}{2} })^{2} } - \sqrt{( {cos \frac{x}{2} - sin \frac{x}{2} })^{2} } } ) = \frac{2}{3} \\ \\ {cot}^{ - 1} ( \frac{cos \frac{x}{2} + sin \frac{x}{2} + cos \frac{x}{2} - sin \frac{x}{2} }{cos \frac{x}{2} + sin \frac{x}{2} - cos \frac{x}{2} + sin \frac{x}{2} } ) = \frac{2}{3} \\ \\ {cot}^{ - 1} ( \frac{2cos \frac{x}{2} }{2sin \frac{x}{2} } ) = \frac{2}{3} \\ \\ {cot}^{ - 1} cot \frac{x}{2} = \frac{2}{3} \\ \\ \frac{x}{2} = \frac{2}{3} \\ \\ x = \frac{4}{3}$