Question

sin a. tana/ 1 - cos a = 1 + sec a

Answer 1

hope it will help you !!

Answer 2

Hi...☺️

Here is your answer...✌️

LHS

$= \frac{ \sin( \alpha ) \tan( \alpha ) }{1 - \cos( \alpha ) } \\ \\ = \frac{ \sin( \alpha ) \times \frac{ \sin( \alpha ) }{ \cos( \alpha ) } }{1 - \cos( \alpha ) } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:( Since \: \tan( \alpha ) = \frac{ \sin( \alpha )}{ \cos( \alpha ) }) \\ = \frac{ \sin {}^{2} ( \alpha ) }{ \cos( \alpha )(1 - \cos( \alpha )) } \\ \\ = \frac{1 - \cos {}^{2} ( \alpha ) }{ \cos( \alpha )( 1 - \cos( \alpha )) } \\ \\ = \frac{(1 + \cos( \alpha ))(1 - \cos( \alpha )) }{ \cos( \alpha )(1 - \cos( \alpha ) ) } \\ \\ = \frac{1 + \cos( \alpha ) }{ \cos( \alpha ) } \\ \\ = \frac{1}{ \cos( \alpha ) } + \frac{ \cos( \alpha ) }{ \cos( \alpha ) } \\ \\ = \sec( \alpha ) + 1$

= RHS [ PROVED ]