How can I prove the trig identity sin2x = 2tanx/(tan^2x+1)?

[ku1005]

the follwing trig identity is used a lot...but i am unable to proove it because i don't know wat to start with in solving it...

sin2x=2tanx/(tan^2x+1)

it invlves the double angle formulae i know that obviously, i tried substituting sin and cos into the RHS of equation and get my answer down to:

2sinxcox/(1/cos^2x) which is close...i think since sin2x is 2sinxcosx...

does any1 have any ideas? LOL...have any ideas...u already know
appreciate any hints
thanks

 

[ku1005]

damm...evry time i post sumthin on here...i look again at the question and get the answer..soz...

i ended up with
(2tanxcos^2x)/(sin^2x+cos^2x) where abviously the bottom = 1 and at the top the cos x's cancel to leave

2sinxcosx = sin2x

thanks though