Question

prove that. sin theta - cos theta +1 / sin theta + cos theta - 1 = 1 / sec theta - tan theta??

Answer 1

hence proved its very easy

Answer 2

HELLO DEAR,

$\frac{ \sin( \alpha ) - \cos( \alpha ) + 1 }{ \sin( \alpha ) + \cos( \alpha ) - 1 } \\ = > \frac{ \cos \alpha (\sin \alpha - \cos( \alpha ) + 1 )}{ \cos \alpha (\sin( \alpha ) + \cos( \alpha ) - 1) } \\ = > \frac{\cos \alpha (\sin \alpha - \cos( \alpha ) + 1 )}{ \cos\alpha \times \sin\alpha + {cos}^{2} \alpha - \cos \alpha } \\ = > \frac{\cos \alpha (\sin \alpha - \cos( \alpha ) + 1 )}{ \cos \alpha \sin\alpha - \cos\alpha + (1 - {sin}^{2} \alpha) } \\ = > \frac{\cos \alpha (\sin \alpha - \cos( \alpha ) + 1 )}{ (1 - \sin \alpha )(1 + \sin \alpha ) - \cos \alpha (1 - \sin \alpha ) } \\ = > \frac{\cos \alpha (\sin \alpha - \cos( \alpha ) + 1 )}{(1 - \sin \alpha) (1 + \sin \alpha - \cos \alpha ) } \\ = > \frac{ \cos \alpha }{(1 - sin \alpha )} \\ = > \frac{ \frac{ \cos \alpha }{ \cos \alpha } }{ \frac{1 - \sin \alpha }{ \cos \alpha } } \: \: \: [DIVIDE IN \: BOTH \: NUMERATOR \: AND \: DENOMINATOR \: by \: \cos \alpha]\\ = > \frac{1}{ \frac{1}{ \cos\alpha } - \frac{ \sin \alpha }{ \cos \alpha } } \\ = > \frac{1}{ \sec \alpha - \tan \alpha }$
$I HOPE ITS HELP YOU DEAR, \\ THANKS$
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