prove sin(A+B)/sin(A-B)=(tanA+tanB)/(tanA-tanB)

Answer 1

tan(A+B) = (tanA+tanB)/(1-tanAtanB)
tan(A-B) = (tanA-tanB)/(1+tanAtanB)

so,

tanA+tanB = tan(A+B)(1-tanAtanB)
tanA-tanB = tan(A-B)(1+tanAtanB)

divide the two and you have

tan(A+B)/tan(A-B) * (1-tanAtanB)/(1+tanAtanB)

multiply top and bottom by cosAcosB nad you have

tan(A+B)/tan(A-B) * (cosAcosB-sinAsinB)/(cosAcosB+sinAsinB)

That's just

tan(A+B)/tan(A-B) * cos(A+B)/cos(A-B)

= sin(A+B)/sin(A-B)

Answer 2

very good