Answer
See the step-by-step proof below.
Work Step by Step
Recall $ sin(A-B)=sinAcosB-sinBcosA $ and $ cos(A-B)=cosAcosB+sinAsinB $
We have:
$ tan(A-B)=\frac{sin(A-B)}{cos(A-B)}=\frac{sinAcosB-sinBcosA}{cosAcosB+sinAsinB}$
Divide both the numerator and the denominator by $ cosAcosB $:
$ tan(A+B)=\frac{tanA-tanB}{1+tanAtanB}$
