If $\frac{\pi }{2}<\theta <\pi$, then $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}$ is equal to is

Solution:

We have,

$\begin{array}{l}\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}\\ =\frac{1-\sin \theta +1+\sin \theta }{\sqrt{1-{\sin }^2\theta }}\\ =\frac{2}{|\cos \theta |}=\frac{2}{-\cos \theta }=-2\sec \theta , \left[\begin{array}{l}\because \pi /2<\theta <\pi \\ \therefore \cos \theta <0\end{array}\right]\end{array}$