Math - Trig - Double Angles

Prove:
sin2x / 1 - cos2x = cotx

My Attempt:
LS:
= 2sinxcosx / - 1 - (1 - 2sin^2x)
= 2sinxcosx / - 1 + 2sin^2x
= cosx / sinx - 1
= cosx / sinx - 1/1
= cosx / sinx - sinx / sinx

--

Prove:
2sin(x+y)sin(x-y) = cos2y - cos2x

My Attempt:
RS:
= 1 - 2sin^2y - 1 - 2sin^2x
= 1 - 1 - 2sin^2y - 2sin^2x
= -2sin^2y - 2sin^2x

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  1. Solved the first problem, I know what I did wrong...

    LS:
    = 2sinxcosx /1 - (1 - 2sin^2x)
    = 2sinxcosx / 1 - 1 + 2sin^2x
    = 2sinxcosx / 2sin^2x
    = cosx / sinx
    = cotx

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  2. sin2x / 1 - cos2x = cotx

    2sinxcosx / 1 - (1-2sin^2x)
    2sinxcosx / 2sin^2x
    cosx/sinx = cotx

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  3. I know how to solve the first question.

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  4. 2sin(x+y)sin(x-y) = cos2y - cos2x

    lhs
    2(sinx cosy + cosx siny) (sin x cos y – cosx siny)
    2( sin^2xcos^2y – sinxcosycosxsiny + sinxcosycosxsiny – cos^2xsin^2y)
    2(sin^2xcos^2y – cos^2xsin^2y)
    2[(1-cos^2x)cos^2y – (1-cos^2y)cos^2x]
    2[cos^2y-cos^2xcos^2y – cos^2x + cos^2xcos^2y]
    2[cos^2y-cos^2x]

    rhs
    2 cos^2y - 1 - 2cos^2x+1
    2[cos^2y – cos^2x]

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  5. thanks

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