Question

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

Answer 1

do I solve it like this:
\[(\sin ^{2}x)(1+\tan ^{2x}) \]
\[\sin ^{2x}\div1\times(sinx \div cosx)\]
\[\sin ^{2x}cosx \div sinx\]

Answer 2

x is suppose to be down, oops.

Answer 3

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

Answer 4

http://www.wolframalpha.com/input/?i=is+%281-cos2x%29%281%2Btan2x%29+equal+to+tan2x

Answer 5

false? .. oh.

Answer 6

how would I know when it's false on a test?

Answer 7

1# wolfram says so
2# it is giving solution ... this is only true in certain points. not on every ponits

Answer 8

i don't understand..

Answer 9

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

Answer 10

sin(x) = cos(pi/2 - x) <---- put any value of x, you will always find it true!!

Answer 11

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

Answer 12

LOL .. i found it a little while ago

Answer 13

Why not just expand the left side and do some cancelling?

Answer 14

that's what I tried but then there's going to be no sinx at the bottom?

Answer 15

It is equivalent to .. proving
\( \-cos 2x ( 1 +\tan2x) - 1 = 0\)

\( \implies \cos2x + \sin2x + 1 = 0\) <--- which is a false identity!!

Answer 16

\( (\tan2x + 1) - (\tan2x+1)\cos2x = \tan 2x \)

\( \implies 1 - (\tan2x+1)\cos2x = 0 \)