Question

1+sina/1-sina = (sec+tana)^2​

Answer 1

$Answer \: \\ \\ GIVEN \: \: Question \: \: Is \: \: \\ \frac{1 + \sin(x) }{1 - \sin(x) } = ( \sec(x) + \tan(x) ) {}^{2} \\ \\ rhs \\ \\ ( \sec(x) + \tan(x) ) {}^{2} \\ \\ ( \frac{1}{ \cos(x) } + \frac{ \sin(x) }{ \cos(x) } ) {}^{2} \\ \\ \frac{(1 + \sin(x) ) {}^{2} }{ \cos {}^{2} (x) } \\ \\ \frac{(1 + \sin(x)) {}^{2} }{1 {}^{2} - \sin {}^{2} (x) } \\ \\ \: \frac{(1 + \sin(x)) \times (1 + \sin(x) )}{(1 - \sin(x)) \times (1 + \sin(x) ) } \\ \\ \frac{(1 + \sin(x) )}{(1 + \sin(x)) } \times \frac{(1 + \sin(x) )}{(1 - \sin(x)) } \\ \\ \frac{1 + \sin(x) }{1 - \sin(x) } \: \: \: \: \: hence \: proved \: \\ \\ Note \: \: \\ \\ 1) \: \: \: \: \sec(x) = \frac{1}{ \cos(x) } \\ \\ 2) \: \: \: \tan(x) = \frac{ \sin(x) }{ \cos(x) } \\ \\ 3) \: \: \: \cos {}^{2} (x) = 1 - \sin {}^{2} (x) \\ \\ 4) \: \: \: (1 - \sin {}^{2} (x ) ) = (1 - \sin(x) ) \times (1 + \sin(x) )$

Answer 2

Step-by-step explanation:

$\begin{lgathered} \\ \frac{1 + \sin(x) }{1 - \sin(x) } = ( \sec(x) + \tan(x) ) {}^{2} \\ \\ RHS \\ \\ ( \sec(x) + \tan(x) ) {}^{2} \\ \\ ( \frac{1}{ \cos(x) } + \frac{ \sin(x) }{ \cos(x) } ) {}^{2} \\ \\ \frac{(1 + \sin(x) ) {}^{2} }{ \cos {}^{2} (x) } \\ \\ \frac{(1 + \sin(x)) {}^{2} }{1 {}^{2} - \sin {}^{2} (x) } \\ \\ \: \frac{(1 + \sin(x)) \times (1 + \sin(x) )}{(1 - \sin(x)) \times (1 + \sin(x) ) } \\ \\ \frac{(1 + \sin(x) )}{(1 + \sin(x)) } \times \frac{(1 + \sin(x) )}{(1 - \sin(x)) } \\ \\ \frac{1 + \sin(x) }{1 - \sin(x) } \: \: \: \: \: \\ \\Hence \: Proved \: \\ \\ \end{lgathered}$