Question

$\frac{ \cot( \alpha ) + \tan( \beta ) }{ \cot( \beta ) + \tan( \alpha ) } = \cot( \alpha ) \tan( \beta )$
prove it

please​

Answer 1

Step-by-step explanation:

$\large \frac{ \cot( \alpha ) + \tan( \beta ) }{ \cot( \beta ) + \tan( \alpha ) } = \cot( \alpha ) \tan( \beta )$

$\small Taking \: L.H.S ,$

$→\large \frac{ \cot( \alpha ) + \tan( \beta ) }{ \frac{1}{ \tan( \beta ) } + \frac{1}{ \cot( \alpha ) } }$

$→\large \frac{ \cot( \alpha ) + \tan( \beta ) }{ \frac{ \cot( \alpha ) + \tan( \beta ) }{ \tan( \beta ) \cot( \alpha ) }}$

$→\large \frac{ \cancel {\cot( \alpha ) + \tan( \beta ) } }{ { 1}{ }} \times \frac{ \cot( \alpha ) \tan( \beta ) } { \cancel{ \cot( \alpha ) + \tan( \beta ) } }$

$→\large \green{ \cot( \alpha ) \tan( \beta ) }$

$→\small \green{ \cot( \alpha ) \tan( \beta ) } = \: R.H.S, hence \: \: proved$