Question

(secA -tanA)^2 =1-sinA/1+sinA


prove the following​

Answer 1

The answer is in the image below. Hope it will help u...

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If yes please mark it as brainlist

Answer 2

$( { \sec( \alpha ) - \tan( \alpha )) }^{2} = \frac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) }$

LHS

$(\sec( \alpha ) - \tan( \alpha ) ) {}^{2} \\ = >( \frac{1}{ \cos( \alpha ) } - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } ) {}^{2} \\ = > ( \frac{1 - \sin( \alpha ) }{ \cos( \alpha ) } ) {}^{2} \\ = > \frac{ {(1 - \sin( \alpha ) )}^{2} }{ { \cos {}^{2} ( \alpha ) } } \\ = > \frac{(1 - \sin( \alpha ) ) {}^{2} }{1 - { \sin {}^{2} ( \alpha ) } } \\ = > \frac{(1 - \sin( \alpha )) {}^{2} }{(1 + \sin( \alpha ) )\times(1 - \sin( \alpha )) } \\ here \: 1 - \sin( \alpha ) cancel \: out \: \\ = \frac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) }$

Hence LHS =RHS

FORMULA USED:

●sec@=1/cos@

●tan@=sin@/cos@

●cos^2@=1-sin^2@

●(a+b)(a-b)=a^2-b^2

PROVED