A = B + C, tanB/tanC = x/y
Starting with (x-y)/(x+y)*sinA,
(x/y - 1)/(x/y+1)*sinA
(tanB/tanC - 1)/(tanB/tanC + 1)*sinA
(tanB - tanC)/(tanB + tanC)*sinA
now multiply top and bottom by cosB.cosC,
(sinB.cosC - sinC.cosB)/(sinB.cosC + sinC.cosB)*sin(B+C)
sin(B-C)/sin(B+C)*sin(B+C)
= sin(B-C)