## Thomas' Calculus 13th Edition

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 $cosAcosB$: $tan(A+B)=\frac{tanA-tanB}{1+tanAtanB}$