Given that $\sin \theta + 2 \cos \theta = 1$,prove that $2 \sin \theta - \cos \theta = 2$.

(N/A) Given,$\sin \theta + 2 \cos \theta = 1$.
Squaring both sides,we get:
$(\sin \theta + 2 \cos \theta)^2 = 1^2$
$\sin^2 \theta + 4 \cos^2 \theta + 4 \sin \theta \cos \theta = 1$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$ and $\cos^2 \theta = 1 - \sin^2 \theta$:
$(1 - \cos^2 \theta) + 4(1 - \sin^2 \theta) + 4 \sin \theta \cos \theta = 1$
$1 - \cos^2 \theta + 4 - 4 \sin^2 \theta + 4 \sin \theta \cos \theta = 1$
$5 - (\cos^2 \theta + 4 \sin^2 \theta - 4 \sin \theta \cos \theta) = 1$
Rearranging the terms:
$4 \sin^2 \theta + \cos^2 \theta - 4 \sin \theta \cos \theta = 5 - 1$
$4 \sin^2 \theta + \cos^2 \theta - 4 \sin \theta \cos \theta = 4$
Recognizing the algebraic identity $(a - b)^2 = a^2 + b^2 - 2ab$,where $a = 2 \sin \theta$ and $b = \cos \theta$:
$(2 \sin \theta - \cos \theta)^2 = 4$
Taking the square root on both sides:
$2 \sin \theta - \cos \theta = 2$
Hence proved.