Question

1-sin theta upon 1+sin theta = (sec theta -tan theta ) hole square

Answer 1

Answer:

$lhs = \frac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) } \\ multiplying \: nr \: and \: dr \: by \: 1 - \sin( \alpha ) \\ = \frac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) } \times \frac{1 - \sin( \alpha ) }{1 - \sin( \alpha ) } \\ = \frac{ {(1 - \sin( \alpha ) )}^{2} }{1 - { \sin }^{2} \alpha } = \frac{1 - { \sin }^{2} \alpha +2 \sin( \alpha ) }{ { \cos}^{2} \alpha } \\ = { \sec}^{2} \alpha - { \tan}^{2} \alpha - 2 \sec( \alpha ) \tan( \alpha ) \\ = ({ \sec( \alpha ) - \tan( \alpha ) })^{2} \\ = rhs$