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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Advanced 2011 · Paper 1 · Mathematics · Trigonometric Equations

Let P=\ θ: θ- θ= 2 θ\ and Q=\ θ: θ+ θ= 2 θ\ be two sets. Then,

Let P={θ:sin⁡θ−cos⁡θ=2cos⁡θ} and Q={θ:sin⁡θ+cos⁡θ=2sin⁡θ}P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\} \text{ and }Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\} be two sets. Then,

Options

  1. AP⊂QP \subset Q and Q−P≠ΦQ-P \neq \Phi
  2. BQ⊄PQ \not \subset P
  3. CP⊄QP \not \subset Q
  4. DP=QP=Q

Correct answer

D. P=QP=Q

Step-by-step solution

P={θ:sin⁡θ−cos⁡θ=2cos⁡θ⇒cos⁡θ(2+1)=sin⁡θ⇒tan⁡θ=2+1Q={θ:sin⁡θ+cos⁡θ=2sin⁡θ⇒sin⁡θ(2−1)=cos⁡θ⇒tan⁡θ=12−1×2+12+1=(2+1)∴P=Q\begin{aligned}& P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta \\ & \Rightarrow \quad \cos \theta(\sqrt{2}+1)=\sin \theta\\ & \Rightarrow \quad \tan \theta=\sqrt{2}+1 \\ & Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta \\ & \Rightarrow \quad \sin \theta(\sqrt{2}-1)=\cos \theta\\ & \Rightarrow \quad \tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}\\ & =(\sqrt{2}+1) \\ & \therefore \quad P=Q \\ & \end{aligned}
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