Let $P = \{θ : \sinθ - \cosθ = \sqrt{2} \cosθ\}$ and $Q = \{θ : \sinθ + \cosθ = \sqrt{2} \sinθ\}$ be two sets. Then

Let P = {θ : sinθ - cosθ = √(2) cosθ} and Q = {θ : sinθ + cosθ = √(2) sinθ} be two sets. Then $\text{Given: } P = \{\theta : \sin\theta - \cos\theta = \sqrt{2}\cos\theta\}$ $\text{Given: } Q = \{\theta : \sin\theta + \cos\theta = \sqrt{2}\sin\theta\}$ $\text{To find: the relation between } P \text{ and } Q$ $\text{(i) Solve } P \text{ for } \tan\theta$ $\text{(ii) Solve } Q \text{ for } \tan\theta$ $\text{(iii) Compare the two sets}$ $\sin\theta = (1+\sqrt{2})\cos\theta$ $\cos\theta = 0 \Rightarrow \sin\theta = 0 \Rightarrow \sin^2\theta + \cos^2\theta = 0$ $\text{contradiction — so } \cos\theta \neq 0$ $\tan\theta = 1+\sqrt{2}$ $\cos\theta = (\sqrt{2}-1)\sin\theta$ $\sin\theta \neq 0$ $\cot\theta = \sqrt{2}-1$ $\tan\theta = \dfrac{1}{\sqrt{2}-1} = \sqrt{2}+1$ $P = \{\theta : \tan\theta = 1+\sqrt{2}\}$ $Q = \{\theta : \tan\theta = 1+\sqrt{2}\}$ $\theta = n\pi + \tan^{-1}\left(1+\sqrt{2}\right),\quad n \in \mathbb{Z}$ $P = Q$ $P \subseteq Q \text{ and } Q \subseteq P \text{ — both true}$ $\text{but neither is a proper subset, so (1), (2), (3) fail}$