If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$? If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$ ?
$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)5$
First I tried plugging in some values for $\theta$ like $0,\frac{\pi}4,\frac{\pi}3,...$ but neither of these known angles worked. But by doing it I realized that for $\theta=\frac{\pi}4+k\pi\quad$, $\tan^2\theta+4\tan\theta=5\quad$ and $2\sin\theta+\cos\theta\neq\sqrt3$ Hence the fourth choice is wrong.
Also tried to expanding,
$$\tan^2\theta+4\tan\theta=\dfrac{\sin^2\theta}{\cos^2\theta}+\dfrac{4\sin\theta}{\cos\theta}=\dfrac{\sin^2\theta+4\sin\theta\cos\theta}{\cos^2\theta}$$But can't continue even writing $4\sin\theta\cos\theta=2\sin2\theta$ doesn't help.

Alternative:
We have $$2\sin\theta=\sqrt3 -\cos \theta$$
Squaring, $$4-4\cos^2 \theta=3+\cos^2\theta-2\sqrt 3 \cos\theta$$
So, $$5\cos^2 \theta-2\sqrt 3 \cos\theta-1=0$$
Or $$\cos \theta=\frac {2\sqrt 3 \pm 4\sqrt 2}{10}$$
Since further conditions haven't been given, both values would be valid, and from here $\tan \theta$ can be calculated.
