Question

prove that 1 + sin/
1 - sin =
(tan + sec )2​

Answer 1

Step-by-step explanation:

we have,

$\frac{1 + \sin( \alpha ) }{1 - \sin( \alpha ) } = \frac{(1 + \sin( \alpha ))^{2} }{(1 - \sin( \alpha )) \times (1 + \sin( \alpha )) } = \frac{(1 + \sin( \alpha ))^{2} }{1 - { (\sin( \alpha ) )}^{2} }$

$= \frac{(1 + \sin( \alpha ))^{2} }{ { (\cos( \alpha ) )}^{2} }$

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