Question

prove please. it's very hard to me

Answer 1

Hope this will help you....

Answer 2

Hey


Given that :-

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


Now ,

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


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


$\frac{ {( \sin( \alpha ) + 1) }^{2} - { \cos( \alpha ) }^{2} }{ { (\sin( \alpha ) + \cos( \alpha )) }^{2} - 1}$


$\frac{ { \sin( \alpha ) }^{2} + 2 \sin( \alpha ) + 1 - { \cos( \alpha ) }^{2} }{ { \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} + 2 \sin(\alpha )\cos( \alpha ) - 1 }$

$\frac{ { \sin( \alpha ) }^{2} - (1 - { \sin( \alpha ) }^{2} ) + 1 + 2 \sin( \alpha ) }{1 - 1 + 2 \sin( \alpha ) \cos( \alpha ) }$


$\frac{ { \sin( \alpha ) }^{2} - 1 + { \sin( \alpha ) }^{2} + 1 + 2 \sin( \alpha ) }{2 \sin( \alpha ) \cos( \alpha ) }$


$\frac{ {2 \sin( \alpha ) }^{2} + 2 \sin( \alpha ) }{2 \sin( \alpha ) \cos( \alpha ) }$


$\frac{2 \sin( \alpha )(1 + \sin( \alpha )) }{2 \sin( \alpha ) \cos( \alpha ) }$

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


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


$\sec( \alpha ) + \tan( \alpha )$


Now ,

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

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


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


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


$= \sec( \alpha ) + \tan( \alpha )$
So

LHS = RHS

♦ Proved ♦

thanks :)

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