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

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(C) For set $P$,we have $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$. Rearranging gives $\sin 	heta = (\sqrt{2} + 1) \cos 	heta$. Multiplying

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$P = Q$

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(C) For set $P$,we have $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$. Rearranging gives $\sin 	heta = (\sqrt{2} + 1) \cos 	heta$. Multiplying by $(\sqrt{2} - 1)$,we get $(\sqrt{2} - 1) \sin 	heta = (\sqrt{2} - 1)(\sqrt{2} + 1) \cos 	heta$,which simplifies to $(\sqrt{2} - 1) \sin 	heta = \cos 	heta$. Thus,$\sin 	heta = \cos 	heta + \sqrt{2} \sin 	heta - \sin 	heta$,which leads to $\sin 	heta + \cos 	heta = \sqrt{2} \sin 	heta$. This is the defining condition for set $Q$. Similarly,starting from $Q$,we can derive the condition for $P$. Therefore,$P = Q$.

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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