If sinθ + 2 cosθ = 1,

Given, $\quad \sin \theta+2 \cos \theta=1$

On squaring both sides, we get

$(\sin \theta+2 \cos \theta)^{2}=1$

$\Rightarrow \quad \sin ^{2} \theta+4 \cos ^{2} \theta+4 \sin \theta \cdot \cos \theta=1$

$\Rightarrow\left(1-\cos ^{2} \theta\right)+4\left(1-\sin ^{2} \theta\right)+4 \sin \theta \cdot \cos \theta=1$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$\Rightarrow \quad-\cos ^{2} \theta-4 \sin ^{2} \theta+4 \sin \theta \cdot \cos \theta=-4$

$\Rightarrow \quad 4 \sin ^{2} \theta+\cos ^{2} \theta-4 \sin \theta \cdot \cos \theta=4$

$\Rightarrow \quad(2 \sin \theta-\cos \theta)^{2}=4 \quad\left[\because a^{2}+b^{2}-2 a b=(a-b)^{2}\right]$

$\Rightarrow \quad 2 \sin \theta-\cos \theta=2$

Hence proved.