Let's derive it. We know that
sin(A+B) = sin(A)cos(B) + sin(B)cos(A)
cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
tan(A+B) = (sin(A+B))/(cos(A+B))
=(sin(A)cos(B) + sin(B)cos(A))/(cos(A)cos(B) - sin(A)sin(B))
Take cos(A)cos(B) out of denominator as common factor:
=(sin(A)cos(B) + sin(B)cos(A))/(cos(A)cos(B)(1 - tan(A)tan(B)))
We can split this up into two separate terms by splitting up the numerator. This means we can cancel out the cosines on the numerator.
=(sin(A)cancel(cos(B)))/(cos(A)cancel(cos(B))(1 - tan(A)tan(B))
+ (sin(B)cancel(cos(A)))/(cancel(cos(A))cos(B)(1 - tan(A)tan(B))
Because tanphi = (sinphi)/(cosphi) this leaves:
(tan(A))/(1 - tan(A)tan(B)) + (tan(B))/(1 - tan(A)tan(B))
Add together again to obtain
tan(A+B) = (tan(A) + tan(B))/(1 - tan(A)tan(B))
