Question

If $cosec\theta =\frac{p+q}{p-q}$, then $\cot \left(\frac{\pi }{4}+\frac{\theta }{2}\right)=$

• A. $\sqrt{\frac{p}{q}}$ or - $\sqrt{\frac{p}{q}}$
• B. $\sqrt{\frac{q}{p}}$ or - $\sqrt{\frac{q}{p}}$
• C. $\sqrt{pq}$ or $\sqrt{pq}$
• D. pq or - pq

Answer 1

Answer: (B)

Given $cosec\theta =\frac{p+q}{p-q}\Rightarrow \frac{1}{\sin \theta }=\frac{p+q}{p-q},$
$\frac{1+\sin \theta }{1-\sin \theta }=\frac{p+q+p-q}{p+q-p+q}$
$\Rightarrow {\left\{\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}\right\}}^2=\frac{p}{q}$
$\Rightarrow {\left\{\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}\right\}}^2=\frac{p}{q}$
$\Rightarrow {\tan }^2\left(\frac{\pi }{4}+\frac{\theta }{2}\right)=\frac{p}{q}$
$\Rightarrow {\cot }^2\left(\frac{\pi }{4}+\frac{\theta }{2}\right)=\frac{q}{p}$