Verify sinx-siny/cosx-cosy=-cot((x+y)/2)

Sin(A+B)=sinAcosB+cosAsinB
sin(A-B)=sinAcosB-cosAsinB
sin(A+B)-sin(A-B)=2cosAsinB
So, if x=A+B and y=A-B, A=(x+y)/2, B=(x-y)/2
sinx-siny=2cosAsinB=2cos((x+y)/2)sin((x-y)/2)

cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
cos(A+B)-cos(A-B)=-2sinAsinB
cosx-cosy=-2sinAsinB=-2sin((x+y)/2)sin((x-y)/2).

So, (sinx-siny)/(cosx-cosy)=-cot((x+y)/2) QED

Do the inverse operation