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