Note that $\tan(A +B) = \frac{\tan A + \tan B}{1-\tan A\tan B}$ holds only if both $(i)\space A\ne \frac{\pi}{2}+k\pi$ and $(ii)\space B\ne \frac{\pi}{2}+l\pi$ for all $k,l\in\mathbb{Z}$ are satisfied. Otherwise, at least one of the values $\tan A$ or $\tan B$ does not exist since $\tan x=\frac{\sin x}{\cos x}$ whilst the zeros of cosine are of the aforementioned form.
The inconvenient case, though, one could simplify directly. Suppose $(i)$ holds, then $ \tan(A+B)= \tan\left(\frac{\pi}{2}+k\pi+B\right)\stackrel{\text{periodicity}}{=}\tan\left(\frac{\pi}{2}+B \right) \stackrel{\text{trig. identity}}{=}-\cot(B).$
Assured of ability to handle the tangent of sum when $(i)$ or $(ii)$ is true, one can assume that both $\cos A$ and $\cos B$ are not $0$ and divide by it in order to obtain the discussed expression.