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    Can't seem to find the markscheme (official or unofficial anywhere ;/!)
    Can anyone help me on this question?

    [sin2x]/[1 + cos2x] = tanx

    I chose to start with the LHS and so far have got;

    [2sinxcosx]/[1 + cosx]

    and now not sure what to do :confused:

    Thanks
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    I take it that 1+cosx is a typo and you meant [2sinxcosx]/[1 + cos2x], so then you can replace cos2x with cos²x-sin²x which is a double angle formula i think, So then you have [2sinxcosx]/[1 + cos²x-sin²x] and as: cos²x=1-sin²x you can sub this in to get [2sinxcosx]/[2cos²x] then cancel out a cosx and the 2s and you got tanx.
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    \frac{\sin 2x}{1+\cos2x}

    \frac{2\sin x\cos x}{1+2\cos^2 x - 1}

    \frac{\sin x}{\cos x}

    \tan x
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    cos2x = (cosx)^2 - (sinx)^2
    1 = (sinx)^2 + (cosx)^2

    when you add them together:


    (sinx)^2 + (cosx)^2 + (cosx)^2 - (sinx)^2 = 2(cosx)^2

    Now we are left with:

    (sin2x)/2(cosx)^2

    We know sin2x = 2sinxcosx

    2sinxcosx/2(cosx)^2

    Hence you are left with sinx/cosx
 
 
 
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