Question

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Prove that (secθ + tanθ - 1)/ (tanθ - secθ + 1) = cosθ/(1 - sinθ)

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Answer 1

HELLO DEAR,

let (θ) theta=alpha

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

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