Question

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

Answer 1

$RHS=\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}\sqrt{\frac{1-\cos (\frac{\pi }{2}+\theta )}{1+\cos (\frac{\pi }{2}+\theta )}}=\sqrt{\frac{2{\sin }^2(\frac{\pi }{4}+\frac{\theta }{2})}{2{\cos }^2(\frac{\pi }{4}+\frac{\theta }{2})}}=\sqrt{{\tan }^2(\frac{\pi }{4}+\frac{\theta }{2})}=\tan (\frac{\pi }{4}+\frac{\theta }{2})=LHS$

Hence Proved