Question

Prove that $\frac{\sqrt{1+\sin \theta }}{\sqrt{1-\sin \theta }}$ = $\tan (\frac{\pi }{4}+\frac{\theta }{2})$

Answer 1

$R.H.S.$

$\tan (\frac{\pi }{4}+\frac{\theta }{2})$

$=\frac{\tan \frac{\pi }{4}+\tan \frac{\theta }{2}}{1-\tan \frac{\pi }{4}.\tan \frac{\theta }{2}}$

$=\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}$

$=\frac{1+\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}}{1-\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}}$

$=\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}$

$=\frac{\sqrt{{(\cos \frac{\theta }{2}+\sin \frac{\theta }{2})}^2}}{\sqrt{{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2})}^2}}$

$=\frac{\sqrt{{\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}+2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{\sqrt{{\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}$

$=\frac{\sqrt{1+\sin \theta }}{\sqrt{1-\sin \theta }}$

$R.H.S.$

Hence proved.