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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.