Question

how to prove this? ​

Answer 1

Here ,

LHS=

$( \frac{1 + sin \alpha - cos \alpha }{1 + sin \alpha + cos \alpha } ) ^{2} \\ \\ = (\frac{1 - cos \alpha + sin \alpha }{1 + cos \alpha + sin \alpha } ) ^{2} \\ \\ we \: know \: that \: \: \: \: \: \: 1 - cosx \: = 2 {sin}^{2} \frac{x}{2} \: \\ and \: \: \: 1 + cosx \: = 2 {cos}^{2} \frac{x}{2} \: \: \: \: and \: \: \: \: sinx \: = 2sin \frac{x}{2} cos \frac{x}{2} \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ................................(1) \\ \\ so \: now \\ \\ = ( \frac{2 {sin}^{2} \frac{ \alpha }{2} + 2sin \frac{ \alpha }{2}cos \frac{ \alpha }{2} }{2 {cos}^{2} \frac{ \alpha }{2} + 2sin \frac{ \alpha }{2}cos \frac{ \alpha }{2} } )^{2} \\ \\ = [ \frac{2sin \frac{ \alpha }{2}(sin \frac{ \alpha }{2} + cos \frac{ \alpha }{2} )}{2cos \frac{ \alpha }{2}( sin \frac{ \alpha }{2} + cos \frac{ \alpha }{2} )} ] ^{2} \\ \\ = ( \frac{sin \frac{ \alpha }{2} }{cos \frac{ \alpha }{2} } ) ^{2} \\ \\ = \frac{ {sin}^{2} \frac{ \alpha }{2} }{ {cos}^{2} \frac{ \alpha }{2} } \\ \\ now \: as \: per \: equation \: (1) \\ \\ = \frac{( \frac{1 - cos \alpha }{2} )}{( \frac{1 + cos \alpha }{2} )} \\ \\ = \frac{1 - cos \alpha }{1 + cos \alpha }$

= RHS

hope it helps......

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