Let $P=\{θ: \sin θ-\cos θ= \sqrt{2} \cos θ\}$ and, $Q=\{θ: \sin θ + \cos θ=\sqrt{2} \sin θ\}$ be two sets. Then,
Answer & explanation
Correct answer: option 4
$θ∈P$
$⇔\sin θ-\cos θ=\sqrt{2} \cos θ$
$⇔(\sin θ-\cos θ)^2=2 \cos^2 θ$
$⇔\sin^2θ+ \cos^2 θ-2 \sin θ \cos θ=2 \cos^2 θ$
$⇔\cos^2 θ + 2 \sin θ \cos θ+ \sin^2θ = 2 \sin^2 θ$
$⇔(\cos θ + \sin θ)^2 = 2 \sin^2 θ$
$⇔\cos θ+ \sin θ = \sqrt{2} \sin θ$
$⇔θ∈Q$
Hence, $P=Q$