Question

(secx - tanx)² = 1 - sinx / 1+sinx

Answer 1

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