Question

Answer Fast please ​

Answer 1

Hi

Hope you understood

Thank you!!

Answer 2

Step-by-step explanation:

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

WE know that :

$\tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \\ \cot( \alpha ) = \frac{ \cos( \alpha ) }{ \sin( \alpha ) }$

$taking \: lhs \\ \frac{ \cos( \alpha ) }{1 - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } + \frac{ \sin( \alpha ) }{1 - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } } \\ \frac{ \cos( \alpha ) }{ \frac{ \cos\alpha - \sin( \alpha ) }{ \cos( \alpha ) } } + \frac{ \sin( \alpha ) }{ \frac{ \sin( \alpha ) - \cos( \alpha ) }{ \sin( \alpha ) } } \\ \frac{ { \cos( \alpha ) }^{2} }{ \cos( \alpha ) - \sin( \alpha ) } + \frac{ { \sin( \alpha ) }^{2} }{ \sin( \alpha ) - \cos( \alpha ) } \\ taking \cos( \alpha ) - \sin( \alpha ) \\ \frac{ { \cos( \alpha ) }^{2} }{ \cos( \alpha ) - \sin( \alpha ) } - \frac{ { \sin( \alpha ) }^{2} }{ \cos( \alpha ) - \sin( \alpha ) } \\ \frac{ { \cos( \alpha ) }^{2} - { \sin( \alpha ) }^{2} }{ \cos( \alpha ) - \sin( \alpha ) } \\ \frac{ (\cos \alpha - \sin\alpha) ( \cos \alpha + \sin\alpha ) }{ \cos( \alpha ) - \sin( \alpha ) } \\ cancelling \\ \cos( \alpha ) - \sin( \alpha ) \\ lhs = \cos( \alpha ) + \sin( \alpha )$

LHS = RHS

HENCE PROVED