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

Given:

 \( \sin \theta+2 \cos \theta=1 \)

To do:

We have to prove that  \( 2 \sin \theta-\cos \theta=2 . \)

Solution:  

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

Squaring on both sides, we get,
$(\sin \theta+2 \cos \theta)^{2}=1$
$\Rightarrow \sin ^{2} \theta+4 \cos ^{2} \theta+4 \sin \theta \cos \theta=1$
$\Rightarrow (1-\cos ^{2} \theta)+4(1-\sin ^{2} \theta)+4 \sin \theta \cos \theta=1$        [Since $\sin ^{2} \theta+\cos ^{2} \theta=1$]

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

$\Rightarrow  4 \sin ^{2} \theta+\cos ^{2} \theta-4 \sin \theta \cos \theta=4$
$\Rightarrow (2 \sin \theta-\cos \theta)^{2}=4$     [Since $a^{2}+b^{2}-2 ab=(a-b)^{2}$]  

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