# 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.

• Add $\frac{\cos^2\theta}{\cos^2\theta}$ to the fraction. Square $2\sin\theta+\cos\theta$. Does that help? Jun 19, 2021 at 19:00
• @DonThousand Yes I got it now, thanks! Jun 19, 2021 at 19:15
• I was wondering if it's possible to get to $\tan(\theta)+2=\sqrt{6}$ as an alternative solution, but I cannot see a clear way to get to that, other than what's have already been proposed...
– zwim
Jun 19, 2021 at 19:40

$$(2\sin\theta+\cos\theta)^2 = 3$$

$$4\sin^2\theta + \cos^2\theta + 4 \sin\theta \cos\theta = 3 \sin^2\theta + 3\cos^2\theta$$

$$\sin^2\theta + 4\sin\theta \cos\theta = 2\cos^2\theta$$

Now dividing both sides by $$\cos^2\theta$$,

$$\tan^2\theta + 4\tan\theta = 2$$

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.

Simply square the constraint and you get $$4 \sin^2(x)+4\sin(x)\cos(x)+\cos^2(x)=3$$ then double angle and trig identity $$3\sin^2(x)+2\sin(2x)=2$$ furthermore this is equivalent to $$2\sin(2x) = 3\cos^2(x)-1$$

and with your definition of the left hand side

$$\tan^2(x)+4\tan(x)=\frac{\sin^2(x)+2\sin(2x)}{\cos^2(x)}$$

we can insert the definition of $$2\sin(2x)$$ we got from the constraint and we get

$$\tan^2(x)+4\tan(x)=\frac{\sin^2(x)+3\cos^2(x)-1}{\cos^2(x)}=\frac{2\cos^2(x)}{\cos^2(x)}=2$$

I hope that helps and I hope I didn't make a mistake ;)