If cosθ + sinθ = $$\sqrt 2 $$ cosθ, then cosθ - sinθ is?

Solution (By Examveda Team)

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