Question

1+secA/secA=(sinA)^2/1-cosA​

Answer 1

Answer:

Hi...mate your question is...

$\frac{1 + \sec( \beta ) }{ \sec( \beta ) } = ( \sin( \beta ) ) ^{2} \div 1 - \cos( \beta ) \\ lhs = \frac{1 + \sec( \beta ) }{ \sec( \beta ) } \\ = \frac{1 + \frac{1}{ \cos( \beta ) } }{ \frac{1}{ \cos( \beta ) } } \\ = \frac{ \frac{ \cos( \beta) +( 1) }{ \cos( \beta ) } }{ \frac{1}{ \cos( \beta ) } } \\ = \frac{ \cos( \beta ) + 1 }{ \cos( \beta ) } \times \cos( \beta ) \\ = \cos( \beta ) + 1$

sorry...mate I couldn't answer your question...

hope...its help you...