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.
Best Answer
$(2\sin\theta+\cos\theta)^2 = 3$
$4\sin^2\theta + \cos^2\theta + 4 \sin\theta \cos\theta = 3 \sin^2\theta + 3\cos^2\theta$
$\sin^2\theta + 4\sin\theta \cos\theta = 2\cos^2\theta$
Now dividing both sides by $\cos^2\theta$,
$\tan^2\theta + 4\tan\theta = 2$