How would I prove the following

sin (x + y) / sin (x - y) = (tan x + tan y) / (tan x - tan y)

Thank you!

sin (x + y) / sin (x - y) =
(sinxcosy+cosxsiny) / sinxcosy-cosxsiny = divide by cosxcosy
(sinx/cosx+siny/cosy) / (sinx/cosx-siny/cosy)=
(tanx+tany) / (tanx-tany)

sin (x + y) / sin (x - y) = (tan x + tan y) / (tan x - tan y)

I'll work on the right-hand side, hoping to reach the left-hand side:

(tan x + tan y) / (tan x - tan y) = [sin(x)/cos(x) + sin(y)/cos(y)] / [sin(x)/cos(x) - sin(y)/cos(y)] =

= {[sin(x)cos(y) + sin(y)cos(x)] / [cos(x)cos(y)]} / {[sin(x)cos(y) - sin(y)cos(x)] / [cos(x)cos(y)]}

cos(x)cos(y) is dividing the numerator and the denominator, so it gets canceled out:

[sin(x)cos(y) + sin(y)cos(x)] / [sin(x)cos(y) - sin(y)cos(x)] =

= sin (x + y) / sin (x - y) = left-hand side.

So the identity has been proven. I hope this helps...