A) \[\sqrt{\frac{p}{q}}\]
B) \[\sqrt{\frac{q}{p}}\]
C) \[\sqrt{pq}\]
D) \[pq\]
Correct Answer: B
Solution :
\[\cos ec\theta =\frac{p+q}{p-q},\sin \theta =\frac{p-q}{p+q}\] \[\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{\left( \frac{p-q}{p+q} \right)}^{2}}}=\frac{2\sqrt{pq}}{(p+q)}\]\[\left| \cot \left( \frac{\pi }{4}+\frac{\theta }{2} \right) \right|=\left| \frac{\cot \frac{\pi }{4}\cot \frac{\theta }{2}-1}{\cot \frac{\pi }{4}+\cot \frac{\theta }{2}} \right|=\left| \frac{\cot \frac{\theta }{2}-1}{\cot \frac{\theta }{2}+1} \right|\] \[=\left| \frac{\cot \frac{\theta }{2}-sin\frac{\theta }{2}}{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}} \right|\] On rationalizing denominator, we get \[\left| \left( \frac{\cos \frac{\theta }{2}-sin\frac{\theta }{2}}{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}} \right)\left( \frac{\cos \frac{\theta }{2}+sin\frac{\theta }{2}}{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}} \right) \right|\] \[=\left| \frac{\cos \theta }{{{\sin }^{2}}\frac{\theta }{2}+co{{s}^{2}}+2\sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right|\] \[=\left| \frac{\cos \theta }{1+\sin \theta } \right|=\left| \frac{2\sqrt{pq}/(p+q)}{1+\frac{(p-q)}{p+q}} \right|=\frac{\sqrt{pq}}{p}=\sqrt{\frac{q}{p}}\]
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