Question

solved this answers the correct answer is 1/secA-tanA​

Answer 1

Answer:

$\frac{ \sin( \alpha ) - \cos( \alpha ) + 1}{ \sin( \alpha ) + \cos( \alpha ) - 1 }$

Divide by cos(α) on both num. and den.

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

re-arrage them,

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

use sec^2(α)-tan^2(α)=1

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

a^2-b^2=(a+b)(a-b)

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

taking common out from denominator,

$= \frac{ \tan( \alpha ) \: + \: \sec( \alpha ) - 1 }{( \sec( \alpha ) - \tan( \alpha ))( \sec( \alpha ) + \: \tan( \alpha ) - 1) }$

cancel the common terms,

$\frac{1}{ \sec( \alpha ) - \tan( \alpha ) }$

hence proved

hope helps you!!!!!!!