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. $\mathrm{P} \subset \mathrm{Q}$ and $\mathrm{Q}-\mathrm{P} \neq \phi$
  2. $\mathrm{Q} \not \subset \mathrm{P}$
  3. $\mathrm{P} \not \subset \mathrm{Q}$
  4. $\quad \mathrm{P}=\mathrm{Q}$

Solution

$\begin{aligned} & \sin \theta-\cos \theta=\sqrt{2} \cos \theta \\ & \Rightarrow \sin \theta=(\sqrt{2}+1) \cos \theta \\ & \Rightarrow \frac{\sin \theta}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\cos \theta\end{aligned}$ $\begin{aligned} & \quad \Rightarrow \frac{\sin \theta(\sqrt{2}-1)}{(\sqrt{2})^2-1^2}=\cos \theta \\ & \Rightarrow \frac{\sin \theta(\sqrt{2}-1)}{2-1}=\cos \theta \\ & \Rightarrow \quad \sqrt{2} \sin \theta-\sin \theta=\cos \theta \\ & \Rightarrow \\ & \Rightarrow \sin \theta+\cos \theta=\sqrt{2} \sin \theta \\ & \therefore \quad\end{aligned}$ $P=Q$

Asked in: MHT CET 2024 (16 May Shift 1)

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