A) \[2\sec \theta \]
B) \[-2\sec \theta \]
C) \[2\text{cosec}\theta \]
D) None of these
Correct Answer: B
Solution :
\[\sqrt{\left( \frac{1-\sin \theta }{1+\sin \theta } \right)}+\sqrt{\left( \frac{1+\sin \theta }{1-\sin \theta } \right)}\] 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 q lying in 2nd quadrant). So the required positive value\[=\frac{-2}{\cos \theta }=-2\,\sec \theta ,\,\left( \frac{\pi }{2}<\theta <\pi \right)\].
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