Take L.H.S sin2A+sin2B/sin2A-sin2B
= sin2A+sin2B/sin2A-sin2B
Put
[sinC+sinD = 2sin(C+D)/2cos(C-D)/2]
[sinC-sinD = 2cos(C+D)/2.sin(C-D)/2]
= 2 sin(2A+2B)/2 cos(2A-2B)/2 / 2 cos(2A+2B) sin(2A-2B)
= sin(A+B).cos(A-B)/cos(A+B).sin(A-B)
= sin(A+B)/cos(A+B) . cos(A-B)/sin(A-B)
= tan(A+B).cot(A-B)
= tan(A+B).1/tan(A-B)
= tan(A+B)/tan(A-B)
∴ Hence we proved sin2A+sin2B/sin2A-sin2B=tan(A+B)/tan(A-B)
