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milliex51Best ResponseYou've already chosen the best response.0
do I solve it like this: \[(\sin ^{2}x)(1+\tan ^{2x}) \] \[\sin ^{2x}\div1\times(sinx \div cosx)\] \[\sin ^{2x}cosx \div sinx\]
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
x is suppose to be down, oops.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
Are you sure that is true: http://www.wolframalpha.com/input/?i=+%28%281cos+2x%29%281%2Btan+2x%29%29+%3D%3D+tan+2x
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=is+%281cos2x%29%281%2Btan2x%29+equal+to+tan2x
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
how would I know when it's false on a test?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
1# wolfram says so 2# it is giving solution ... this is only true in certain points. not on every ponits
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
i don't understand..
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=+%28%281cos+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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
sin(x) = cos(pi/2  x) < put any value of x, you will always find it true!!
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
LOL .. i found it a little while ago
 one year ago

TransendentialPIBest ResponseYou've already chosen the best response.0
Why not just expand the left side and do some cancelling?
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
that's what I tried but then there's going to be no sinx at the bottom?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
It is equivalent to .. proving \( \cos 2x ( 1 +\tan2x)  1 = 0\) \( \implies \cos2x + \sin2x + 1 = 0\) < which is a false identity!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\( (\tan2x + 1)  (\tan2x+1)\cos2x = \tan 2x \) \( \implies 1  (\tan2x+1)\cos2x = 0 \)
 one year ago
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