Question

prove that 1+sin teta/1-sin teta=(sec teta+tan teta)whole square

Answer 1

$\frac{1 + \sin \theta }{1 - \sin \theta} = \frac{(1 + \sin \theta)(1 + \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} \\ \frac{ {(1 + \sin \theta)}^{2} }{1 - { \sin}^{2} \theta} = \frac{ {(1 + \sin \theta)}^{2} }{ { \cos}^{2} \theta} \\ {( \frac{1 + \sin \theta}{ \cos \theta } )}^{2} = {( \sec \theta + \tan \theta) }^{2}$