Question

tan-1 (√1+x-√1-x/√1+x+√1-x)= π/4-1/2.

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Answer 1

${tan}^{ - 1} ( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } ) \\ \\ put \: \: \: \: x = cos2 \theta \\ \\ 2 \theta = {cos}^{ - 1} x \\ \\ \theta = \frac{1}{2} {cos}^{ - 1}x \\ \\ {tan}^{ - 1} ( \frac{ \sqrt{1 + cos2 \theta} - \sqrt{1 - cos 2\theta} }{ \sqrt{1 + cos2 \theta} + \sqrt{1 - cos \theta} } ) \\ \\ = {tan}^{ - 1} ( \frac{ \sqrt{2 {cos}^{2} \theta } - \sqrt{ 2{sin}^{2} \theta} }{ \sqrt{2 {cos}^{2} \theta} + \sqrt{2 {sin}^{2} \theta } } ) \\ \\ = {tan}^{ - 1} ( \frac{cos \theta - sin \theta}{cos \theta + sin \theta} ) \\ \\ = {tan}^{ - 1} ( \frac{1 - tan \theta}{1 + tan \theta} ) \\ \\ = {tan}^{ - 1} tan( \frac{\pi}{4} - \theta) \\ \\ = \frac{\pi}{4} - \theta \\ \\ = \frac{\pi}{4} - \frac{1}{2} {cos}^{ - 1} x$