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If $\operatorname{cosec} \theta=\frac{p+q}{p-q}$, then $\cot \left(\frac{\pi}{4}+\frac{\theta}{4}\right)$ is equal to
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The correct answer is:
$\frac{\sqrt{q}}{p}$
We have,
$\begin{array}{r}\frac{1}{\sin \theta}=\frac{p+q}{p-q} \\ \frac{1+\tan ^2 \theta / 2}{2 \tan \theta / 2}=\frac{p+q}{p-q}\end{array}$
Applying componendo and dividendo,
$\begin{aligned} & \frac{1+\tan ^2 \theta / 2+2 \tan \theta / 2}{1+\tan ^2 \theta / 2-2 \tan \theta / 2}=\frac{p+q+p-q}{p+q-p+q} \\ & \Rightarrow \quad \frac{(1+\tan \theta / 2)^2}{(1-\tan \theta / 2)^2}=\frac{2 p}{2 q}=\frac{p}{q} \\ & \Rightarrow \quad \frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}=\sqrt{\frac{p}{q}} \\ & \Rightarrow \quad \tan \left(\frac{\theta}{2}+\frac{\pi}{4}\right)=\sqrt{\frac{p}{q}} \\ & \Rightarrow \quad \cot \left(\frac{p}{4}+\frac{\theta}{2}\right)=\sqrt{\frac{p}{q}} \\ & \end{aligned}$
$\begin{array}{r}\frac{1}{\sin \theta}=\frac{p+q}{p-q} \\ \frac{1+\tan ^2 \theta / 2}{2 \tan \theta / 2}=\frac{p+q}{p-q}\end{array}$
Applying componendo and dividendo,
$\begin{aligned} & \frac{1+\tan ^2 \theta / 2+2 \tan \theta / 2}{1+\tan ^2 \theta / 2-2 \tan \theta / 2}=\frac{p+q+p-q}{p+q-p+q} \\ & \Rightarrow \quad \frac{(1+\tan \theta / 2)^2}{(1-\tan \theta / 2)^2}=\frac{2 p}{2 q}=\frac{p}{q} \\ & \Rightarrow \quad \frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}=\sqrt{\frac{p}{q}} \\ & \Rightarrow \quad \tan \left(\frac{\theta}{2}+\frac{\pi}{4}\right)=\sqrt{\frac{p}{q}} \\ & \Rightarrow \quad \cot \left(\frac{p}{4}+\frac{\theta}{2}\right)=\sqrt{\frac{p}{q}} \\ & \end{aligned}$
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