Question

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

Answer 1

We know that

$\sin \theta =\frac{x}{\sqrt{1+x^2}}$

$\cos \theta =\frac{1}{\sqrt{1+x^2}}$

A/Q, $\sin \theta +2\cos \theta =1$

$\Rightarrow \frac{x+2}{\sqrt{1+x^2}}=1$

$\Rightarrow x+2=\sqrt{1+x^2}$

$\Rightarrow \sin \theta =\frac{x}{x+2},\cos \theta =\frac{1}{x+2}$

Now, $2\sin \theta -\cos \theta =\frac{2x-1}{x+2}$

Now, $x+2=\sqrt{1+x^2}\Rightarrow x^2+4x+4=1+x^2$

$\Rightarrow 4x=-3\Rightarrow x=-3/4$

$\therefore$ $2\sin \theta -\cos \theta =\frac{\frac{-3}{2}-1}{\frac{-3}{4}+2}=\frac{-5/2}{-5/4}=4/2=2$

Hence, proved.