Answer: A
Given: \[ \cos\theta + \sin\theta = \sqrt{2} \cos\theta \] ⇒ \(\sqrt{2} \cos\theta = \cos\theta + \sin\theta\) ⇒ \((\sqrt{2} \cos\theta)^2 = (\cos\theta + \sin\theta)^2\) ⇒ \(2 \cos^2\theta = \cos^2\theta + \sin^2\theta + 2 \sin\theta \cos\theta\) ⇒ \(2 \cos^2\theta - \cos^2\theta - 2 \sin\theta \cos\theta = \sin^2\theta\) ⇒ \(\cos^2\theta - 2 \sin\theta \cos\theta = \sin^2\theta\) ⇒ \(\sin^2\theta + \cos^2\theta - 2 \sin\theta \cos\theta = \sin^2\theta + \sin^2\theta\) ⇒ \((\sin\theta - \cos\theta)^2 = 2 \sin^2\theta\) ⇒ \(\sin\theta - \cos\theta = \sqrt{2 \sin^2\theta} = \sqrt{2} \sin\theta\)
Given: \[ \cos\theta + \sin\theta = \sqrt{2} \cos\theta \] ⇒ \(\sqrt{2} \cos\theta = \cos\theta + \sin\theta\) ⇒ \((\sqrt{2} \cos\theta)^2 = (\cos\theta + \sin\theta)^2\) ⇒ \(2 \cos^2\theta = \cos^2\theta + \sin^2\theta + 2 \sin\theta \cos\theta\) ⇒ \(2 \cos^2\theta - \cos^2\theta - 2 \sin\theta \cos\theta = \sin^2\theta\) ⇒ \(\cos^2\theta - 2 \sin\theta \cos\theta = \sin^2\theta\) ⇒ \(\sin^2\theta + \cos^2\theta - 2 \sin\theta \cos\theta = \sin^2\theta + \sin^2\theta\) ⇒ \((\sin\theta - \cos\theta)^2 = 2 \sin^2\theta\) ⇒ \(\sin\theta - \cos\theta = \sqrt{2 \sin^2\theta} = \sqrt{2} \sin\theta\)