How is it that $$\tan(A +B) = \frac{\tan A + \tan B}{1-\tan A\tan B}$$ for any value of $A$, $B$?
I have doubts about this since we arrive at this by dividing the numerator and denominator of $$\frac{\sin(A+B)}{\cos(A+B)}$$ by $\cos A \cdot \cos B$, which can only be done when $\cos A\cdot\cos B$ is not equal to zero.