Question

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

Whether the given equation $2\sin \theta -\cos \theta =2$ is ?

• A. True
• B. False

Answer 1

The correct option is B False

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

Squaring both sides,

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

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

$\Rightarrow 3{\cos }^2\theta +4\sin \theta \cos \theta =0$

$\Rightarrow 3{\cos }^2\theta =-4\sin \theta \cos \theta$

$\Rightarrow 3\cos \theta =-4\sin \theta$ or $\cos \theta =0$

$\Rightarrow \tan \theta =\frac{-3}{4},\therefore \tan \theta$ is in second or fourth quadrant.

since $4^2+3^2=5^2$

Hypotenuse$=5$units

$Case:1:$

$\Rightarrow \sin \theta =\frac{4}{5},\cos \theta =\frac{-3}{5}$ when $\tan \theta$ in second quadrant.

$\therefore 2\sin \theta -\cos \theta =2\frac{4}{5}-\frac{-3}{5}=\frac{8+3}{5}=\frac{11}{5}$

$Case2$

$\Rightarrow \sin \theta =\frac{-4}{5},\cos \theta =\frac{3}{5}$ when $\tan \theta$ in fourth quadrant.

$\therefore 2\sin \theta -\cos \theta =2\frac{-4}{5}-\frac{3}{5}=\frac{-8+3}{5}=\frac{-51}{5}=1$

$Case:3:$

$\cos \theta =0$ from above

$\Rightarrow \theta =(2n+1)\frac{\pi }{2}$

$\Rightarrow \sin \theta =n\pi$

Hence in any of three cases we observe that $2\sin \theta -\cos \theta =2$ is false.