A-B in the summation and product should be replaced by a different notation such as x_i-x_j where underscore means sub(script), and i and j are triangular. If 1, 2, 3 are the vertices of a triangle ABC in clockwise progression, so that y is the adjacent vertex to x clockwise, then i=1 denotes A, i=2 denotes B and i=3 denotes C, and respectively j=2, denoting B, j=3, denoting C, and j=1, denoting A.
So we have x₁-x₂, x₂-x₃, x₃-x₁ representing A-B, B-C, C-A respectively as i goes from 1 to 3.
PROOF
tan(A-B)=(x-y)/(1+xy), tan(B-C)=(y-z)/(1+yz), tan(C-A)=(z-x)/(1+xz).
∑tan(x_i-x_j)=
[(x-y)(1+yz)(1+xz)+(y-z)(1+xy)(1+xz)+(z-x)(1+xy)(1+yz)]/((1+xy)(1+yz)(1+xz))=
[(x-y)(1+xz+yz+xyz²)+(y-z)(1+xy+xz+x²yz)+(z-x)(1+xy+yz+xy²z)]/((1+xy)(1+yz)(1+xz))=
[x+x²z+x²yz²-y-y²z-xy²z²+y+xy²+x²y²z-z-xz²-x²yz²+z+yz²+xy²z²-x-x²y-x²y²z]/((1+xy)(1+yz)(1+xz))=
[x²z-y²z+xy²-xz²+yz²-x²y]/((1+xy)(1+yz)(1+xz)).
ᴨtan(x_i-x_j)=(x-y)(y-z)(z-x)/((1+xy)(1+yz)(1+xz)).
(x-y)(yz-xy-z²+xz)=x²z-y²z+xy²-xz²+yz²-x²y.
Therefore
∑tan(x_i-x_j)=ᴨtan(x_i-x_j)