Question
Given that $\sin\theta+2\cos\theta=1,$ then prove that $2\sin\theta-\cos\theta=2.$
✓
Answer
$\sin\theta+2\cos\theta=1\ [\text{Given}]$
On squaring both sides, we get
$(\sin\theta)^2+(2\cos\theta)^2+2(\sin\theta)(2\cos\theta)=1$
$\Rightarrow\ \sin^2\theta+4\cos^2\theta+4\sin\theta\cos\theta=1$
$\Rightarrow\ 1-\cos^2\theta+4(1-\sin^2\theta)+4\sin\theta\cos\theta=1$
$\Rightarrow\ 1-\cos^2\theta+4-4\sin^2\theta+4\sin\theta\cos\theta=1$
$\Rightarrow\ -\cos^2\theta-4\sin^2\theta+4\sin\theta\cos\theta=-4$
$\Rightarrow\ \cos^2\theta+4\sin^2\theta-4\sin\theta\cos\theta=4$
$\Rightarrow\ (\cos\theta)^2+(2\sin\theta)^2-2(\cos\theta)(2\sin\theta)=4$
$\Rightarrow\ (2\sin\theta-\cos\theta)^2=2^2$
Taking square root both sides, we have
$2\sin\theta-\cos\theta=2$
Hence, proved.
On squaring both sides, we get
$(\sin\theta)^2+(2\cos\theta)^2+2(\sin\theta)(2\cos\theta)=1$
$\Rightarrow\ \sin^2\theta+4\cos^2\theta+4\sin\theta\cos\theta=1$
$\Rightarrow\ 1-\cos^2\theta+4(1-\sin^2\theta)+4\sin\theta\cos\theta=1$
$\Rightarrow\ 1-\cos^2\theta+4-4\sin^2\theta+4\sin\theta\cos\theta=1$
$\Rightarrow\ -\cos^2\theta-4\sin^2\theta+4\sin\theta\cos\theta=-4$
$\Rightarrow\ \cos^2\theta+4\sin^2\theta-4\sin\theta\cos\theta=4$
$\Rightarrow\ (\cos\theta)^2+(2\sin\theta)^2-2(\cos\theta)(2\sin\theta)=4$
$\Rightarrow\ (2\sin\theta-\cos\theta)^2=2^2$
Taking square root both sides, we have
$2\sin\theta-\cos\theta=2$
Hence, proved.
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