(b) $\sqrt {\left( {\frac{{1 – \sin \theta }}{{1 + \sin \theta }}} \right)} + \sqrt {\left( {\frac{{1 + \sin \theta }}{{1 – \sin \theta }}} \right)} $

is the sum of two positive quantities and hence the result must be positive. 

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

we have the sum equal to $\frac{{1 – \sin \theta + 1 + \sin \theta }}{{\sqrt {1 – {{\sin }^2}\theta } }} = \frac{2}{{\cos \theta }};$ 

which is negative.

( $\because$ $\cos \theta $ is negative for $\theta$ lying in $2^{nd}$ quadrant). 

So the required positive value $ = \frac{{ – 2}}{{\cos \theta }} = – 2\,\sec \theta ,\,\left( {\frac{\pi }{2} < \theta < \pi } \right)$.