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

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

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$P = Q$

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(C) For set $P$,we have $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$.This implies $\sin 	heta = (\sqrt{2} + 1) \cos 	heta$.Multiplying both sides by $(\sqrt{2} - 1)$,we get $(\sqrt{2} - 1) \sin 	heta = (\sqrt{2} - 1)(\sqrt{2} + 1) \cos 	heta$.$(\sqrt{2} - 1) \sin 	heta = (2 - 1) \cos 	heta = \cos 	heta$.Thus,$\sqrt{2} \sin 	heta - \sin 	heta = \cos 	heta$,which gives $\sqrt{2} \sin 	heta = \sin 	heta + \cos 	heta$.This is the defining condition for set $Q$.Similarly,starting from the condition for set $Q$,$\sin 	heta + \cos 	heta = \sqrt{2} \sin 	heta$,we can derive $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$.Since every element of $P$ is in $Q$ and every element of $Q$ is in $P$,we conclude that $P = Q$.

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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