Question

If $\sin \theta +2\cos \theta =1$
Prove $\cos \theta -2\sin \theta =2$

Answer 1

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

$\Rightarrow {(\sin \theta +2\cos \theta )}^2=1^2$ [ Squaring ]

$\Rightarrow {\sin }^2\theta +4{\cos }^2\theta +4\sin \theta \cos \theta =1$

$\Rightarrow (1-{\cos }^2\theta )+4\sin \theta \cos \theta +4(1-{\sin }^2\theta )=1$

$\Rightarrow 1-{\cos }^2\theta +4\sin \theta \cos \theta +4-4{\sin }^2\theta =1$

$\Rightarrow -4{\sin }^2\theta +4\sin \theta \cos \theta -{\cos }^2\theta =1-1-4$

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

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

$\Rightarrow {(2\sin \theta -\cos \theta )}^2={\left(2\right)}^2$

$\Rightarrow 2\sin \theta -\cos \theta =\pm 2$

$\Rightarrow 2\sin \theta -\cos \theta =2$

Or $2\sin \theta -\cos \theta =-2$

$\Rightarrow \cos \theta -2\sin \theta =2$

Hence, the answer proved.