Well, there are addition formulas for $\tan$:
$\begin{align} \tan (v + u) = \frac{\tan v + \tan u}{1 - \tan v \tan u} \\ \tan (v - u) = \frac{\tan v - \tan u}{1 + \tan v \tan u} \end{align}$
So you can compute $\tan(hdg)$ and $\tan(\theta)$ then plug them into the first formula to get $\tan(\theta+hdg)$.
For the other formula you get have $\tan (90^\circ - w) = \cot w = \frac{1}{\tan w}$ where $w = hdg - \theta$, i.e.
$\tan (90^\circ - (hdg - \theta)) = \frac{1}{\tan (hdg-\theta)} = \frac{1}{\frac{\tan hdg - \tan \theta}{1+\tan hdg \tan \theta}} = \frac{1 + \tan hdg \tan \theta}{\tan hdg - \tan \theta}$
so you can once again use your previously calculated values for $\tan \theta$ and $\tan hdg$.
You might want to look into another way of evaulating $\tan x$. If your angle $x$ is smaller than $22.5^\circ$ you can use the polynomial
$\tan x \approx x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315}$
Note that this is for an angle in radians, so you have to convert your angle from degrees to radians by multiplying by $\frac{\pi}{180} \approx 0.0174533$ You probably want to evaulate this using Horner's rule. Depending on how sensitive things are you might be able to drop one (or two) terms.
If your angle is larger than $22.5^\circ$ (or what limit you find suitable), you can use the half angle formula $\tan x = \frac{2 \tan \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$ or the reciprocal $\tan x = \frac{1}{\tan (90^\circ - x)}$ depending on how large your angle is.
I don't know if this will be quicker or not, but it might be worth a shot.