Question

√1+sinA/1-sinA=secA+tanA​

Answer 1

To prove :

$\sf \star \: \sqrt{ \frac{1 + \sin( \theta) }{1 - \sin( \theta)} } = \sec( \theta) + \tan( \theta)$

Proof :

$\sf \hookrightarrow \sqrt{ \frac{1 + \sin( \theta) }{1 - \sin( \theta)} } \\ \\ \sf \hookrightarrow \frac{ \sqrt{1 + \sin( \theta)} }{ \sqrt{1 - \sin( \theta) } } \times \frac{ \sqrt{1 + \sin( \theta)} }{ \sqrt{1 + \sin( \theta) } } \\ \\ \sf \hookrightarrow \frac{ ({ \sqrt{(1 + \sin( \theta)} \: )}^{2} }{ \sqrt{ {(1)}^{2} - { \sin }^{2}( \theta) } } \\ \\ \sf \hookrightarrow \frac{1 + \sin( \theta) }{ \sqrt{ {\cos}^{2} ( \theta) } } \\ \\ \sf \hookrightarrow \frac{1 + \sin( \theta) }{ \cos( \theta) } \\ \\ \sf \hookrightarrow\frac{1}{ \cos( \theta) } + \frac{ \sin( \theta) }{ \cos( \theta) } \\ \\ \sf \hookrightarrow \sec( \theta) + \tan( \theta)$

Hence proved

Identities used :

$\sf \star \: \: \frac{1}{ \cos( \theta) } </u><u>= \sec( \theta)</u><u> \\ \\ \sf\star \: \: \frac{ \sin( \theta) }{ \cos( \theta) } = \tan( \theta) \\ \\ \sf\star \: \: 1 - { \sin}^{2}( \theta) = { \cos}^{2} (\theta)$