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# C3 solving problem - stuck, help :(! Watch

1. 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

Thanks
2. 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.

3. cos2x = (cosx)^2 - (sinx)^2
1 = (sinx)^2 + (cosx)^2

(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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