If Sin(A+B)=3 Sin(A-B), Show that tan A=2 tan B

Question

Trigonometry question

Answer 1

sin(A+B) = sinAcosB +sinBcosA

=3sin(A-B) = 3sinAcosB -3sinBcosA = sinAcosB +sinBcosA

2sinAcosB = 4sinBcosA

sinAcosB = 2sinBcosA

sinAcosB/sinBcosA = 2

2 = sinAcosB/sinBcosA = tanAcotB = tanA/tanB

2 = tanA/tanB

tanA = 2tanB