If you have two Pythagorean triples a²+b²=x² and c²+d²=y² you could be said to implicitly be talking about two triangles (a,b,x) and (c,d,y). One of the angles in these two triangles is tan⁻¹(b/a) and in the other is tan⁻¹(d/c). If you put these two angles adjacent to each other and scale the (c,d,y) triangle up to use the (a,b) vector as the basis for its x-axis and a (-b, a) vector as the basis for its y-axis, then its (c, 0) side has become (ca, cb), and the side from (c, 0) to (c, d) now has a displacement of (-bd, ad), so the new corner is at (ca-bd, cb+ad), so the tangent of the angle sum is going to be (cb+ad)/(ca-bd).