Given, sin θ + 2 cos θ = 1\xa0On squaring both sides, we get{tex}(sin\\theta+2cos\\theta)^2=1{/tex}{tex}\\Rightarrow sin^2\\theta+4cos^2\\theta+4sin\\theta cos\\theta=1{/tex}⇒ 1 – cos2\xa0θ + 4 (1 – sin2\xa0θ) + 4 sin θ cos θ = 1\xa0{tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}⇒ 1 – cos2\xa0θ + 4 – 4 sin2\xa0θ + 4 sin θ cos θ = 1⇒ –cos2\xa0θ – 4 sin2\xa0θ + 4 sin θ cos θ = –4{tex}\\Rightarrow -(cos^2\\theta+4sin^2\\theta-4sin\\theta cos\\theta)=-4{/tex}⇒ cos2\xa0θ + 4 sin2\xa0θ – 4 sin θ cos θ = 4⇒ (cos θ)2\xa0+ (2 sin θ)2\xa0– 2(cos θ) (2 sin θ) = 4⇒ (2 sin θ – cos θ)2\xa0= 22{tex}\\Rightarrow{/tex}2 sin θ – cos θ = 2{tex}{/tex}Hence\xa0proved.