Prove that sin2A+sin2B/sin2A-sin2B=tan(A+B)/tan(A-B)

Steps to prove sin2A+sin2B/sin2A-sin2B=tan(A+B)/tan(A-B)

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)

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