Note $A+B+C=\pi$, therefore $$\tan nC = \tan n(\pi - (A+B)) = \tan(n\pi - n(A+B))= -\tan n(A+B)$$ The given to-be-claimed identity is equivalent to $$-\tan nC = \frac{\tan nA + \tan nB}{1-\tan nA \tan nB}$$ Replace $\tan nC = -\tan (nA +nB)$ gives $$\tan (nA + nB) = \frac{\tan nA + \tan nB}{1 - \tan nA \tan nB}$$ The last relation obviously holds by the sum of angle formula. (Note: we have not discussed the case when $(\tan nA \tan nB = 1)$ which will prevent us from dividing $(1-\tan nA \tan nB)$ when transforming the to-be-claimed identity. But this case can be easily handled.)