Solve for $\theta$: $a = b\tan\theta - \frac{c}{\cos\theta}$ This question was initially posted on SO (Link).  I'm not sure the answer given there was correct.  I cannot get the results from those expressions to match my CAD model.
The title pretty much sums it up. How do I solve for theta given the following equation.
$$a = b\tan\theta - \frac{c}{\cos\theta}$$
I am not a student and this is not homework. It's been quite a while since I've done any significant trig and I'm out of time to figure this out.

Using $\tan2A=\dfrac{2t}{1-t^2},\cos2A=\dfrac{1-t^2}{1+t^2},$ where $t=\tan A$
we have $a=b\dfrac{2t}{1-t^2}+c\dfrac{1+t^2}{1-t^2},$ where $t=\tan\dfrac\theta2$
rearrange to form a Quadratic Equation in $t=\tan\dfrac\theta2$
