Question

Prove $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}$ $=\frac{-2}{\cos \theta },\frac{\pi }{2}<\theta <\pi$.

Answer 1

Consider the L.H.S.

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

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

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

$=\frac{2}{\sqrt{1-{\sin }^2\theta }}$

$=\frac{2}{\sqrt{{\cos }^2\theta }}$

$=\frac{2}{\cos \theta }$

Since, for $\frac{\pi }{2}<\theta <\pi$

Therefore,

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

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

Hence, proved.