Pre-calculus

prove the identity
sin2x-sin2y/sin2x+sin2y=tan(x-y)/tan(x+y)

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  1. This one is easier if you work on the right side.

    (sin(x-y)/cos(x-y)) / (sin(x+y)/cos(x+y))

    (sinxcosy-cosxsiny)/(cosxcosy+sinxsiny)
    -------------------------------------
    (sinxcosy+cosxsiny)/(cosxcosy-sinxsiny)

    Expand those products, and you will be able to factor out some (sin^2+cos^2) and watch it become just like the left side.

    Working on the left side first requires you to insert some canceling factors which are not obvious.

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