Question

If $2\sin \theta +\cos \theta =2,0^0\le \theta \le {90}^0,$ then $\sin \theta -2\cos \theta$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

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

$\therefore \cos \theta =2-2\sin \theta$ .....(1)

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

Substituting the value of $\cos \theta$ from (1)

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

$\Rightarrow 5{\sin }^2\theta -8\sin \theta +3=0$

$\Rightarrow (\sin \theta -1)(5\sin \theta -3)=0$ ......(2)

$\therefore$ either $5\sin \theta -3=0$

$\Rightarrow \sin \theta =\frac{3}{5}$

$\Rightarrow \cos \theta =\frac{4}{5}$

Putting these values in the given equation, we get

$\sin \theta -2\cos \theta =\frac{3}{5}-2\times \frac{4}{5}=-1$

We reject this value as

$0^{\circ }\le \theta \le {90}^{\circ }$

Coming to the other factor of equation(2)

$\sin \theta -1=0$

$\Rightarrow \sin \theta =1$

$\Rightarrow \cos \theta =0$

$\Rightarrow \sin \theta -2\cos \theta =1-2\times 0=1$