Question

$sin \theta = \sqrt{ \frac{1-x}{1+x} }\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater$
$Prove \: that \: \: \: \: \: \frac{sec\theta \: + \: tan\theta}{sec\theta \: - \: tan\theta} = \frac{1+ \sqrt{1- {x}^{2} } }{x}$
________________​

Answer 1

Step-by-step explanation:

$here \: \: © =(θ) \\ \\ \frac{sec©+tan©}{sec©-tan©} \\ = \frac{ \frac{1}{cos©}+ \frac{sin©}{cos©} } { \frac{1}{cos©} - \frac{sin©}{cos©} } \\ = \frac{1 + sin©}{1-sin©} \\ = \frac{1 + \sqrt{ \frac{1 - x}{1 + x} } }{1 - \sqrt{ \frac{1 - x}{1 + x} } } \\ = \frac{ \sqrt{1 + x} + \sqrt{1 - x} }{ \sqrt{1 + x} - \sqrt{1 - x} } \\ =( \frac{ \sqrt{1 + x} + \sqrt{1 - x} }{ \sqrt{1 + x} - \sqrt{1 - x} } )( \frac{\sqrt{1 + x} + \sqrt{1 - x} }{\sqrt{1 + x} + \sqrt{1 - x} } ) \\ = \frac{(\sqrt{1 + x} + \sqrt{1 - x} )^{2} }{1 + x - 1 + x} \\ = \frac{1 + x + 2 \sqrt{1 - {x}^{2} } + 1 - x}{1 + x - 1 + x} \\ = \frac{2 + 2 \sqrt{1 - {x}^{2} } }{2x} \\ = \frac{1 + \sqrt{1 - {x}^{2} } }{x}$

Answer 2

Answer:

vsggdgdgd

Step-by-step explanation:

hjsjehejnsbdbebrbrbrbrbrjrj