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Question: If \(\text{cosec}\theta = \frac{p + q}{p - q},\) then \(\cot\left( \frac{\pi}{4} + \frac{\theta}{2} ...

If cosecθ=p+qp−q,\text{cosec}\theta = \frac{p + q}{p - q}, then cot⁡(π4+θ2)=\cot\left( \frac{\pi}{4} + \frac{\theta}{2} \right) =

A

pq\sqrt{\frac{p}{q}}

B

qp\sqrt{\frac{q}{p}}

C

pq\sqrt{pq}

D

pqpq

Answer

qp\sqrt{\frac{q}{p}}

Explanation

Solution

Given, cosecθ=p+qp−q⇒1sin⁡θ=p+qp−q,co\text{sec}\theta = \frac{p + q}{p - q} \Rightarrow \frac{1}{\sin\theta} = \frac{p + q}{p - q},

Apply componendo and dividendo,

1+sin⁡θ1−sin⁡θ=p+q+p−qp+q−p−q⇒[cos⁡θ2+sin⁡θ2cos⁡θ2−sin⁡θ2]2=pq\frac{1 + \sin\theta}{1 - \sin\theta} = \frac{p + q + p - q}{p + q - p - q} \Rightarrow \left\lbrack \frac{\cos\frac{\theta}{2} + \sin\frac{\theta}{2}}{\cos\frac{\theta}{2} - \sin\frac{\theta}{2}} \right\rbrack^{2} = \frac{p}{q}⇒[1+tan⁡θ/21−tan⁡θ/2]2=pq⇒tan⁡2(π4+θ2)=pq⇒cot⁡2(π4+θ2)=qp\left\lbrack \frac{1 + \tan\theta/2}{1 - \tan\theta/2} \right\rbrack^{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}Note : cot⁡(π4+θ2)=qp\cot\left( \frac{\pi}{4} + \frac{\theta}{2} \right) = \sqrt{\frac{q}{p}} only if cot⁡(π4+θ2)>0\cot\left( \frac{\pi}{4} + \frac{\theta}{2} \right) > 0.