Question

Let $P=\{\theta :sin\theta -cos\theta =\sqrt{2}cos\theta \}$ and $Q=\{sin\theta +cos\theta =\sqrt{2}sin\theta \}$ be two sets. Then:

• A. $P\subset QandQ-P\ne \varnothing$
• B. $Q\subset P$
• C. $P\subset Q$
• D. $P=Q$

Answer 1

The correct option is D $P=Q$

$P=\left\{\theta :\sin \theta -\cos \theta =\sqrt{2}\cos \theta \right\}$

$Q=\left\{\theta :\sin \theta +\cos \theta =\sqrt{2}\sin \theta \right\}$

From $P$

$\sin \theta -\cos \theta =\sqrt{2}\cos \theta$

$\sin \theta =\left(\sqrt{2}+1\right)\cos \theta$

$\frac{\sin \theta }{\cos \theta }=\left(\sqrt{2}+1\right)$

$\tan \theta =\left(\sqrt{2}+1\right)$

from $Q$

$\sin \theta +\cos \theta =\sqrt{2}\sin \theta$

$\sin \theta \left(\sqrt{2}-1\right)=\cos \theta$

$\frac{\sin \theta }{\cos \theta }=\left(\sqrt{2}-1\right)$

$\tan \theta =\left(\sqrt{2}-1\right)$

$\tan \theta =\frac{1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$

$\therefore P=Q$

Hence,

option $\left(D\right)$ is correct answer.