Question

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

• A. $\sqrt{q/p}$
• B. $\sqrt{p/q}$
• C. $pq$
• D. $\sqrt{pq}$

Answer 1

The correct option is A $\sqrt{q/p}$

$cosec\theta =\frac{p+q}{p-q}$

$\Rightarrow \frac{1}{\sin \theta }=\frac{p+q}{p-q}$

Applying componendo and dividendo

$\Rightarrow \frac{1+\sin \theta }{1-\sin \theta }=\frac{p+q+p-q}{p+q-p+q}$

$\Rightarrow {\left(\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}\right)}^2=\frac{p}{q}$

$\Rightarrow {\left(\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}\right)}^2=\frac{p}{q}$

$\Rightarrow {\tan }^2(\frac{\pi }{4}+\frac{\theta }{2})=\frac{p}{q}$

$\Rightarrow {\cot }^2(\frac{\pi }{4}+\frac{\theta }{2})=\frac{q}{p}$

$\Rightarrow \cot (\frac{\pi }{4}+\frac{\theta }{2})=\sqrt{\frac{q}{p}}$