Question

$\sqrt{ 1 + \sin( \alpha ) \div 1 - \sin( \alpha )} = \sec( \alpha ) + \tan( \alpha )$
​Please prove this sum

Answer 1

Step-by-step explanation:

We have,

$\sqrt{ \frac{1 + \sin( \alpha ) }{1 - \sin( \alpha ) } }$

$= \sqrt{ \frac{(1 + \sin( \alpha ) )(1 + \sin( \alpha ) )}{(1 - \sin( \alpha ))(1 + \sin( \alpha ) )} }$

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

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

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

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

$= \sec ( \alpha ) + \tan( \alpha )$