TRIG IDENTITY! math help please!?

please prove

[cos4x•tan2x - sin4x] / [cos4x•cot2x + sin4x] = -tan^2(2x)

please show as much work as possible :P thanks so much for your time! I'll select as best answer as soon as yahoo allows me!

Update:

hi Josh, I have reported you. Please don't be rude to others. I've seen your other answers and they are very hurtful. If people are asking for advice, do not hurt them and make their situations worse. Please respect the answers community.

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    hey that was easy than i thought

    i thought it will be difficult :)

    but it's not :)

    finally i did it XD

    solution-

    LHS:

    let do it part by part :)

    numerator-

    [cos4x•tan2x - sin4x]

    (cos4x * sin2x/cos2x - sin4x)

    (cos4x * sin2x - sin4x * cos2x)/cos2x

    use identity

    sin(α – β) = sin(α)cos(β) – cos(α)sin(β)

    sin(2x-4x)/cos2x

    -sin 2x / cos 2x

    -tan 2x

    denominator:

    [cos4x•cot2x + sin4x]

    cos4x * cos 2x/sin2x + sin4x

    cos(α – β) = cos(α)cos(β) + sin(α)sin(β)

    (cos 4x * cos2x + sin 4x * sin 2x)/sin 2x

    cos(4x-2x)/sin2x

    cos2x/sin2x

    =cot2x

    -tan 2x/cot2x

    -tan2x/(1/tan2x)

    -tan^2x <=== answer :)

    Source(s): grade 12 now :)
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