<s>[INST] Prove $\tan(A+B)$ using $\cos(A-B)$ and $\sin(A-B)$ Use $\cos(A-B)$ and $\sin(A-B)$ to prove $$\tan(A+B)=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$$
It seems like we cannot simply change $A+B$ to $A+(-B)$ to prove it? Any ideas?
 [/INST] $$\begin{align*}\tan (A+B) = \frac{\sin (A+B)}{\cos (A+B)} &= \frac{\sin A\cos B + \sin B\cos A}{\cos A\cos B - \sin A\sin B} \div \frac{\cos A\cos B}{\cos A\cos B} \\ & = \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{1 - \frac{\sin A\sin B}{\cos A\cos B}} \\ & = \frac{\tan A + \tan B}{1 - \tan A \tan B}\end{align*}$$

Note that if you know $\sin (A-B)$ then use $B \to -B$ to get $\sin (A+B)$ with the usual odd sine and even cosine, same goes for $\cos(A-B)$. 
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