If cosθ+sinθ=2cosθ, then prove that sinθ-cosθ=2sinθ
Prove the given expression
Given,
cosθ+sinθ=2cosθ⇒cosθ+sinθ2=2cosθ2⇒cos2θ+2cosθsinθ+sin2θ=2cos2θ∵a+b2=a2+2ab+b2⇒2cosθsinθ+sin2θ=2cos2θ-cos2θ⇒2cosθsinθ+sin2θ=cos2θ⇒sin2θ=cos2θ-2cosθsinθ⇒sin2θ+sin2θ=cos2θ+sin2θ-2cosθsinθ[addingsin2θonbothside]⇒2sin2θ=sinθ-cosθ2∵a2-2ab+b2=a-b2⇒2sin2θ=sinθ-cosθ2Takingrootonbothside⇒2sinθ=sinθ-cosθ
Hence Proved, sinθ-cosθ=2sinθ