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

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(D) For set $P$: $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$ $\Rightarrow \sin 	heta = (\sqrt{2} + 1) \cos 	heta$ $\Rightarrow 	an 	heta =

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$P = Q$

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(D) For set $P$: $\sin 	heta - \cos 	heta = \sqrt{2} \cos 	heta$ $\Rightarrow \sin 	heta = (\sqrt{2} + 1) \cos 	heta$ $\Rightarrow 	an 	heta = \sqrt{2} + 1$ For set $Q$: $\sin 	heta + \cos 	heta = \sqrt{2} \sin 	heta$ $\Rightarrow \cos 	heta = (\sqrt{2} - 1) \sin 	heta$ $\Rightarrow \frac{1}{\sqrt{2} - 1} = 	an 	heta$ $\Rightarrow 	an 	heta = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \sqrt{2} + 1$ Since both sets $P$ and $Q$ represent the same set of values for $	heta$ satisfying $	an 	heta = \sqrt{2} + 1$,we have $P = Q$.

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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