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

On squaring both sides, we get

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

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

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

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

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

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

$2 \sin \theta-\cos \theta=2$ Hence proved.