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

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(D) For set $P$,we have $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$.This simplifies to $\sin 	heta = (\sqrt{2} + 1) \cos 	heta$,which implies

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$P = Q$

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(D) For set $P$,we have $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$.This simplifies to $\sin 	heta = (\sqrt{2} + 1) \cos 	heta$,which implies $	an 	heta = \sqrt{2} + 1$.For set $Q$,we have $\sin 	heta + \cos 	heta = \sqrt{2} \sin 	heta$.This simplifies to $\cos 	heta = (\sqrt{2} - 1) \sin 	heta$,which implies $	an 	heta = \frac{1}{\sqrt{2} - 1}$.Rationalizing the denominator,$	an 	heta = \frac{1(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1$.Since both sets $P$ and $Q$ represent the same condition $	an 	heta = \sqrt{2} + 1$,it follows that $P = Q$.Therefore,option $(D)$ is the 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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