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# P2 trig question watch

1. Prove that:

cot2x + cosec2x = cotx

This is what I've done so far, which gives me the wrong answere.

cot2x +cosec2x

(1-tan^2x)/(2tanx) + 1/(2sinxcosx)

[(1-sin^2x/cos^2x) / (2sinx/cosx)] + 1/(2sinxcosx)

(cosx)/(2sinx) + 1/(2sinxcosx)

(cosx + 1)/(2sinx)

(cotx + 1)/2

2. [(1-sin^2x/cos^2x) / (2sinx/cosx)]
=(cosx - (sin^2x/cosx)) / 2sinx

the other fraction shares 2sinx, so add them together and times this one by cos x

along those lines anyway.

what you wrote for that fraction isnt right

it comes to cos^2x - sin^2x + 1 / (2sinxcosx)
which comes out to cot x = cosx/sinx
3. I've included that though, what do you mean?

(to stevo)
4. cot2x = 1/tan2x
cosec2x = 1/sin2x

cos2x/sin2x + 1/sin2x = [cos2x + 1]/sin2x

cos2x = (cos^2)x - (sin^2)x therefore cos2x + 1 = (cos^2)x - (sin^2)x + 1
sin2x = 2sinxcosx

[cos2x+1]/[sin2x] = [(cos^2)x - {(sin^2)x - 1}]/[2sinxcosx]
= [(cos^2)x - {-(cos^2)x}]/[2sinxcosx]
= [2(cos^2)x]/[2sinxcosx]
= (cos^2)x / sinxcosx
= cosx / sinx
= cotx

edit: decided to show whole method...
5. OHHHH!

Thank you. I see it now, that's really annoying.

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