Question

Cos theta / 1 - tan theta + sin theta / 1- cot theta = (cos theta + sin theta )

Answer 1

Hey mate here we go

$lets \: given \: eq \: is \: i \: will \: take \\ theta \: =alpha( \alpha ) \\ \frac{ \cos( \alpha ) }{1 - \tan( \alpha ) } + \frac{ \sin( \alpha ) }{1 - \cot( \alpha ) } \\ lets \: \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \\ \cot( \alpha ) = \frac{ \cos( \alpha ) }{ \sin( \alpha ) } \\ nw \: put \: those \: value \: in \: that \: eq \\ \frac{ \cos( \alpha ) }{1 - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } + \frac{ \sin( \alpha ) }{1 - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } } \\ \frac{ \cos( \alpha ) }{ \frac{ \cos( \alpha) - \sin( \alpha ) }{ \cos( \alpha ) } } + \frac{ \sin( \alpha ) }{ \frac{ \sin( \alpha) - \cos( \alpha ) }{ \sin( \alpha ) } } \\ \frac{ { \cos}^{2} \alpha }{ \cos( \alpha ) - \sin( \alpha ) } + \frac{ { \sin }^{2} \alpha }{ \sin( \alpha ) - \cos( \alpha ) } \\ ( \frac{ { \cos}^{2} \alpha }{ \cos( \alpha ) - \sin( \alpha ) } - \frac{ { \sin }^{2} \alpha }{ \cos( \alpha ) - \sin( \alpha ) } ) \\ now \: ( \frac{ { \cos }^{2} \alpha - { \sin }^{2} \alpha }{ \cos( \alpha ) - \sin( \alpha ) } ) \\ = ( \frac{( \cos( \alpha ) + \sin( \alpha ) )( \cos( \alpha ) - \sin( \alpha )) }{ \cos( \alpha ) - \sin( \alpha ) } \\ now \: - \: value \:will \: cancelled \\ u \: will \: only \: get \: \cos( \alpha ) + \sin( \alpha ) = r \: h \: s$

I hope this will helps you

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