If tanx=2tany, prove that sin(x+y)/sin(x-y)=3

LHS = \frac{\sin (x + y)}{\sin (x - y)} = \frac{\sin x . \cos y + \cos x . \sin y}{\sin x . \cos y - \cos x . \sin y} = \frac{\frac{\sin x . \cos y}{\cos x . \cos y} + \frac{\cos x . \sin y}{\cos x . \cos y}}{\frac{\sin x . \cos y}{\cos x . \cos y} - \frac{\cos x . \sin y}{\cos x . \cos y}} [dividing numerator and denominator by \cos x . \cos y] = \frac{\tan x + \tan y}{\tan x - \tan y} = \frac{2 \tan y + \tan y}{2 \tan y - \tan y} [as \tan x = 2 \tan y, given] = \frac{3 \tan y}{\tan y} = 3 = RHS h e n c e, LHS = RHS

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