help 20

One angle T is divided in two part A & B where tanA/tanB=x/y. Prove that -

sin(A-B)=(x-y)/(x+y)*sinT

sinT  = sin(A + B)

x = tan A,  y  = tan B

 (x - y)/(x + y)sin(A + B)  =  

[tanA - tanB] / [tanA + tanB] sin(A + B)  =

[(sinA/cosA) - (sinB/cosB)] / [ (sinA/cosA) + (sinB/cosB)] sin(A + B) =

[(sinAcosB - sinBcosA) / cosAcosB] / [(sinAcosB + sinBcosA)/ cosAcosB] sin(A + B) =

 ([sinAcosB - sinBcosA] /[ [sinAcosB + sinBcosA] )  *  [ sinAcosB + sinBcosA]  =

[sinAcosB - sinBcosA]  =

sin(A - B)