This is the basic and exam oriented question from trigonometry, topic – trigonometric identities in which we have given that (sinθ+cosθ)=√2sinθ, and we have been asked to show that (sinθ-cosθ) = √2cosθ

RS Aggarwal, Class 10, chapter 13B, question no 12

cosθ+sinθ=√2cosθ

squaring on both the sides, we get,

cos²θ+sin²θ+2sinθcosθ=2cos²θ

cos²θ−sin²θ=2cosθsinθ

(cosθ+sinθ)(cosθ−sinθ)=2cosθsinθ

√2cosθ(cosθ−sinθ) = 2cosθsinθ [Given cosθ+sinθ = √2cosθ]

∴cosθ−sinθ = √2sinθ [henceproved]