#### Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

Solution.  Given:-   sinθ + 2cosθ = 1
squaring both sides we have
$\left ( \sin \theta +2\cos \theta \right )^{2}= 1^{2}$
$\sin ^{2}\theta +4\cos ^{2}\theta +4\sin \theta \cos \theta = 1$
$\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
$\sin ^{2}\theta +4\cos ^{2}\theta = 1-4\sin \theta \cos \theta \cdots \left ( 1 \right )$
To prove :

$2\sin \theta -\cos \theta = 2$
Taking the left-hand side
$2\sin \theta -\cos \theta \cdots \left ( 2 \right )$
On squaring equation (2) we get
$\left (2 \sin \theta -\cos \theta \right )^{2}$
$= 4\sin ^{2}\theta +\cos ^{2}\theta -4\sin \theta \cos \theta$  $\left ( \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right )$
$= 3\sin ^{2}\theta +\sin ^{2}\theta+\cos ^{2}\theta -4\sin \theta \cos \theta$
$= 3\sin ^{2}\theta +1-4\sin \theta \cos \theta$         $\left [ \because \sin ^{2} \theta +\cos ^{2} \theta= 1\right ]$
$= 3\sin ^{2} \theta+\sin^{2} \theta+4\cos ^{2} \theta$
[use equation (1)]
$= 4\sin ^{2}\theta +4\cos ^{2}\theta$
$= 4\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )$
= 4

$\left [ \because \sin ^{2} \theta +\cos ^{2} \theta= 1\right ]$
So here we get the value of (2sin$\theta$ – cos$\theta$)2 is 4
$\left ( 2\sin \theta -\cos \theta \right )^{2}= 4$
$2\sin \theta -\cos \theta = \sqrt{4}$
$2\sin \theta -\cos \theta = 2$
Hence proved