Question

prove that ....inverse trigonometry​

Answer 1

${ \bold{ \green{ \underline{ANSWER} : - }}} \\ \\ \\ { \bold{ \underline{TO \: \: PROVE} : - }} \\ \\ { \bold{ {tan}^{ - 1}( \frac{ \sqrt{1 + {x}^{2} } - \sqrt{1 - {x}^{2} } }{ \sqrt{1 + {x}^{2} } + \sqrt{1 - {x}^{2} } }) = \frac{\pi}{4} - \frac{1}{2} {cos}^{ - 1}( {x}^{2} ) }} \\ \\ { \bold{ \red{ \underline{L.H.S.} : - } }} \\ \\ { \bold{ \pink{ = { \tan}^{ - 1} ( \frac{ \sqrt{1 + {x}^{2} } - \sqrt{1 - {x}^{2} } }{ \sqrt{1 + {x}^{2} } - \sqrt{1 - {x}^{2} } }) }}} \\ \\ { \bold{ \orange{put \: \: {x}^{2} = \cos(2 \alpha ) }}} \\ \\ { \bold{ \pink{ = { \tan}^{ - 1}( \frac{ \sqrt{1 + \cos(2 \alpha ) } - \sqrt{1 - \cos(2 \alpha ) } }{ \sqrt{1 + \cos(2 \alpha ) } + \sqrt{1 - \cos(2 \alpha ) } } } }} \\ \\ { \bold{ \pink{ = {tan}^{ - 1} ( \frac{ \sqrt{2 {cos}^{2}( \alpha ) } - \sqrt{2 { \sin}^{2} ( \alpha )} }{ \sqrt{2 { \cos }^{2}( \alpha ) } + \sqrt{2 { \sin}^{2} \alpha } }) }}} \\ \\ { \bold{ \pink{ = {tan}^{ - 1}( \frac{ \cos( \alpha ) - \sin( \alpha ) }{ \cos( \alpha ) + \sin( \alpha ) } )}}} \\ \\ { \bold{ \pink{ = {tan}^{ - 1} ( \frac{1 - \tan( \alpha ) }{1 + \tan( \alpha ) }) }}} \\ \\ { \bold{ \pink{ = { \tan }^{ - 1} ( \tan( \frac{\pi}{4} - \alpha ) }}} \\ \\ { \bold{ \pink{ = \frac{\pi}{4} - \alpha }}} \\ \\ { \bold{ \pink{ = \frac{\pi}{4} - \frac{1}{2} { \cos }^{ - 1} ( {x}^{2} ) }}} \\ \\ { \bold{ \red{ = R.H.S. \: \: \: (H.P.)}}} \\ \\ \\ \\ { \bold{ \boxed{ \huge{ \red{FOLLOW \: ME...}}}}}$