A) \[\sqrt{\frac{p}{q}}\]
B) \[\sqrt{\frac{q}{p}}\]
C) \[\sqrt{pq}\]
D) \[pq\]
Correct Answer: B
Solution :
Given, \[\text{cosec}\theta =\frac{p+q}{p-q}\] Þ \[\frac{1}{\sin \theta }=\frac{p+q}{p-q}\] Apply componendo and dividendo \[\frac{1+\sin \theta }{1-\sin \theta }=\frac{p+q+p-q}{p+q-p+q}\] Þ \[{{\left\{ \frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}} \right\}}^{2}}=\frac{p}{q}\] Þ \[{{\left\{ \frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}} \right\}}^{2}}=\frac{p}{q}\] Þ \[{{\tan }^{2}}\left( \frac{\pi }{4}+\frac{\theta }{2} \right)=\frac{p}{q}\] Þ \[{{\cot }^{2}}\left( \frac{\pi }{4}+\frac{\theta }{2} \right)=\frac{q}{p}\] Note: \[\cot \left( \frac{\pi }{4}+\frac{\theta }{2} \right)=\sqrt{\frac{q}{p}}\,\text{only,}\,\,\text{if}\]\[\cot \,\left( \frac{\pi }{4}+\frac{\theta }{2} \right)>0\].
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