• # question_answer If $\text{cosec}\theta =\frac{p+q}{p-q},$ then $\cot \,\left( \frac{\pi }{4}+\frac{\theta }{2} \right)=$ [EAMCET 2001] A) $\sqrt{\frac{p}{q}}$ B) $\sqrt{\frac{q}{p}}$ C) $\sqrt{pq}$ D) $pq$

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$.