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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
$P \subset Q$ and $Q-P \neq \varnothing$
$Q \not \subset P$
$P \not \subset Q$
$P=Q$
Solution
$P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}$
$Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}$
From P
$\sin \theta-\cos \theta=\sqrt{2} \cos \theta$
$\sin \theta=(\sqrt{2}+1) \cos \theta$
$\frac{\sin \theta}{\cos \theta}=(\sqrt{2}+1)$
$\tan \theta=(\sqrt{2}+1)$
$\text { from } Q$
$\sin \theta+\cos \theta=\sqrt{2} \sin \theta$
$\sin \theta(\sqrt{2}-1)=\cos \theta$
$\frac{\sin \theta}{\cos \theta}=(\sqrt{2}-1)$
$\tan \theta=(\sqrt{2}-1)$
$\tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$
$\therefore P=Q$
Hence, option $(D)$ is correct answer.