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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

If cosec θ=(p+q)/(p-q) , then ((π)/(4)+(θ)/(4))is equal to

If cosec⁡θ=p+qp−q\operatorname{cosec} \theta=\frac{p+q}{p-q}, then cot⁡(π4+θ4)isequalto\cot \left(\frac{\pi}{4}+\frac{\theta}{4}\right)is equal to

Options

  1. Apq\frac{\sqrt{p}}{q}
  2. Bqp\frac{\sqrt{q}}{p}
  3. Cpq\sqrt{p q}
  4. Dpqp q

Correct answer

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

Step-by-step solution

We have, 1sin⁡θ=p+qp−q1+tan⁡2θ/22tan⁡θ/2=p+qp−q\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, 1+tan⁡2θ/2+2tan⁡θ/21+tan⁡2θ/2−2tan⁡θ/2=p+q+p−qp+q−p+q⇒(1+tan⁡θ/2)2(1−tan⁡θ/2)2=2p2q=pq⇒tan⁡π4+tan⁡θ21−tan⁡π4tan⁡θ2=pq⇒tan⁡(θ2+π4)=pq⇒cot⁡(p4+θ2)=pq\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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