If cosθ + sinθ = 2 cosθ, then cosθ - sinθ is?

Answer: Option D

$$\begin{array}{rl}& \text{cos}\theta +\sin \theta =\sqrt{2}\cos \theta \\ & \text{Squaring both sides}\\ & \text{co}{\text{s}}^2\theta +{\sin }^2\theta +2\cos \theta .\sin \theta =2\text{co}{\text{s}}^2\theta \\ & \Rightarrow 2\text{co}{\text{s}}^2\theta -\text{co}{\text{s}}^2\theta -{\sin }^2\theta =2\text{cos}\theta .\text{sin}\theta \\ & \Rightarrow \text{co}{\text{s}}^2\theta -{\sin }^2\theta =2\sin \theta .\text{cos}\theta \\ & \Rightarrow \left(\cos \theta -\sin \theta \right)\left(\cos \theta +\sin \theta \right)=2\sin \theta .\text{cos}\theta \\ & \Rightarrow \left(\cos \theta -\sin \theta \right)\left(\sqrt{2}\cos \theta \right)=2\sin \theta .\text{cos}\theta \\ & \Rightarrow \cos \theta -\sin \theta =\frac{2\sin \theta .\cos \theta }{\sqrt{2}\cos \theta }\\ & \Rightarrow \sqrt{2}\sin \theta \\ & \\ & \mathbf{Alternate}:\\ & \text{Let }\sqrt{2}\cos \theta =\alpha \\ & \therefore \cos \theta \pm \sin \theta =a\\ & \cos \theta \pm \sin \theta =\sqrt{2-a^2}\\ & =\sqrt{2-a^2}\\ & =\sqrt{2-2\text{co}{\text{s}}^2\theta }\\ & =\sqrt{2\left(1-\text{co}{\text{s}}^2\theta \right)}\\ & =\sqrt{2{\sin }^2\theta }\\ & =\sqrt{2}\sin \theta \end{array}$$