Gujarati
Hindi
1.Set Theory
hard

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

A

$P \subset Q$ and $Q-P \neq \varnothing$

B

$Q \not \subset P$

C

$P \not \subset Q$

D

$P=Q$

(IIT-2011)

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.

Standard 11
Mathematics

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