How can I solve for a single variable which occurs in multiple trigonometric functions in an equation? This is a pretty dumb question, but it's been a while since I had to do math like this and it's escaping me at the moment (actually, I'm not sure I ever knew how to do this. I remember the basic trigonometric identities, but not anything like this).
I have a simple equation of one unknown, but the unknown occurs twice in different trigonometric functions and I'm not sure how to combine the two. I want to simply solve for $\theta$ in the following equation, where $a$ and $b$ are constants.
$a=\tan(\theta) - \frac{b}{\cos^2\theta}$
How can I reduce this into a single expression so that I can solve for $\theta$ given any $a$ and $b$?
(I'm only interested in real solutions and, in practice (this is used to calculate the incidence angle for a projectile such that it will pass through a certain point), it should always have a real solution, but an elegant method of checking that it doesn't would not go unappreciated.)

Based on Braindead's hint I reduced the equation to:
$0=(b-a)+\tan(\theta)+b\tan^2(\theta)$
I can now solve for $\tan(\theta)$ using the quadratic equation, which gets me what I'm after. Is this the solution others were hinting towards? It seems like there would be a way to do it as a single trigonometric operation, but maybe not.
 A: Hint:
Can you solve $$p = \frac{q\sin 2\theta + r}{s\cos 2\theta + t}$$
Ok, more details.
$$a = \frac{\sin \theta \cos \theta}{\cos^2 \theta} - \frac{b}{\cos^2 \theta}  = \frac{\sin 2 \theta }{2\cos^2 \theta} - \frac{b}{\cos^2 \theta} $$
$$ = \frac{\sin 2\theta - 2b}{2cos^2 \theta} = \frac{ \sin 2\theta - 2b}{\cos 2\theta + 1}$$
Thus
$$a(\cos 2 \theta + 1) = \sin 2 \theta - 2 b$$
Thus
$$   \sin 2\theta - a \cos 2\theta = a + 2b$$
The equation
$$ p \cos \alpha + q \sin \alpha = r$$ 
is standard.
and can be solved by dividing by $\displaystyle \sqrt{p^2 + q^2}$ and noticing that for some $\displaystyle \beta$ we must have that $\displaystyle \sin \beta = \frac{p}{\sqrt{p^2 + q^2}}$ and $\displaystyle  \cos \beta = \frac{q}{\sqrt{p^2 + q^2}}$
Giving rise to 
$$ \sin(\alpha + \beta) = \frac{r}{\sqrt{p^2 +q^2}}$$ 
I will leave it to you to solve your original equation.
A: You can write $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\sqrt{1-\cos^2(\theta)}}{\cos(\theta)}$ which gets everything in terms of $\cos(\theta)$ but you may not like the degree of the result when you get rid of the radical.
