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How do I prove (1-cos2x)(1+tan2x)=tan2x?

Mathematics
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do I solve it like this: \[(\sin ^{2}x)(1+\tan ^{2x}) \] \[\sin ^{2x}\div1\times(sinx \div cosx)\] \[\sin ^{2x}cosx \div sinx\]
x is suppose to be down, oops.
Are you sure that is true: http://www.wolframalpha.com/input/?i=+%28%281-cos+2x%29%281%2Btan+2x%29%29+%3D%3D+tan+2x

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Other answers:

http://www.wolframalpha.com/input/?i=is+%281-cos2x%29%281%2Btan2x%29+equal+to+tan2x
false? .. oh.
how would I know when it's false on a test?
1# wolfram says so 2# it is giving solution ... this is only true in certain points. not on every ponits
i don't understand..
http://www.wolframalpha.com/input/?i=+%28%281-cos+2%28pi%2F3%29%29%281%2Btan+2%28pi%2F3%29%29%29+%3D%3D+tan+2%28pi%2F3%29 it is not true for x=pi/3
sin(x) = cos(pi/2 - x) <---- put any value of x, you will always find it true!!
I don't know how this TrueQ works: http://www.wolframalpha.com/input/?i=TrueQ%5Bsin+%5E2+x+%2B+cos+%5E2+x+%3D+1%5D
LOL .. i found it a little while ago
Why not just expand the left side and do some cancelling?
that's what I tried but then there's going to be no sinx at the bottom?
It is equivalent to .. proving \( \-cos 2x ( 1 +\tan2x) - 1 = 0\) \( \implies \cos2x + \sin2x + 1 = 0\) <--- which is a false identity!!
\( (\tan2x + 1) - (\tan2x+1)\cos2x = \tan 2x \) \( \implies 1 - (\tan2x+1)\cos2x = 0 \)

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