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)$.
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