Question

Prove $\sqrt{\frac{1+sinA}{1-sinA} }$ = $\frac{cosA}{1-sinA}$

Answer 1

$\sqrt{ \frac{1 + \sin(a) }{1 - \sin(a) } } = \frac{ \cos(a) }{1 - \sin(a) } \\ lhs = \sqrt{ \frac{1 + \sin(a) }{1 - \sin(a) } } \\ = \frac{ \sqrt{1 + \sin(a) } }{ \sqrt{1 - \sin(a) } } \times \frac{ \sqrt{1 + \sin(a) } }{1 + \sin(a) } \\ = \frac{( { \sqrt{1 + \sin(a)) } }^{2} }{ \sqrt{1 - { \sin(a) }^{2} } } \\ = \frac{1 + \sin(a) }{ \sqrt{ { \cos(a) }^{2} } } \\ = \frac{1 + \sin(a) }{ \cos(a) } \\ \\ rhs = \frac{ \cos(a) }{1 - \sin(a) } \\ = \frac{ \cos(a) }{1 - \sin(a) } \times \frac{1 + \sin(a) }{1 + \sin(a) } \\ = \frac{ \cos(a)(1 + \sin(a)) }{1 - { \sin(a) }^{2} } \\ = \frac{ \cos(a)( 1 + \sin(a)) }{ { \cos(a) }^{2} } \\ = \frac{1 + \sin(a) }{ \cos(a) } \\ \\ \\ hence \: as \: lhs = rhs \\ \\ \\ so \: proved$