Let P={theta: sin theta-cos theta=sqrt{2} cos | vedclass

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Question

$P=\{	heta: \sin 	heta-\cos 	heta=\sqrt{2} \cos 	heta\}$

$Q=\{	heta: \sin 	heta+\cos 	heta=\sqrt{2} \sin 	heta\}$

From P

$\sin 	heta

Vedclass · Accepted Answer

$P=Q$

Vedclass · Answer

$P=\{	heta: \sin 	heta-\cos 	heta=\sqrt{2} \cos 	heta\}$

$Q=\{	heta: \sin 	heta+\cos 	heta=\sqrt{2} \sin 	heta\}$

From P

$\sin 	heta-\cos 	heta=\sqrt{2} \cos 	heta$

$\sin 	heta=(\sqrt{2}+1) \cos 	heta$

$\frac{\sin 	heta}{\cos 	heta}=(\sqrt{2}+1)$

$	an 	heta=(\sqrt{2}+1)$

$	ext { from } Q$

$\sin 	heta+\cos 	heta=\sqrt{2} \sin 	heta$

$\sin 	heta(\sqrt{2}-1)=\cos 	heta$

$\frac{\sin 	heta}{\cos 	heta}=(\sqrt{2}-1)$

$	an 	heta=(\sqrt{2}-1)$

$	an 	heta=\frac{1}{\sqrt{2}-1} 	imes \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$

$	herefore P=Q$

Hence, option $(D)$ is correct answer.

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

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