Let $P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}$ and $Q=\{\theta: \sin \theta+\cos…

Let $P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}$ and $Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}$ be two sets. Then,
  1. $P \subset Q$ and $Q-P \neq \Phi$
  2. $Q \not \subset P$
  3. $P \not \subset Q$
  4. $P=Q$

Solution

$ \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} $

Asked in: JEE Advanced 2011 (Paper 1)

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