Prove the identity (2tanx)/(x+tan^2x) = sin2x

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Prove the identity (2tanx)/(x+tan^2x) = sin2x

Mathematics
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@freckles haha sorry me again..with more trig. I've gotten [(2tanx)/(1-tan^2x)][cos^2x]
Ah no, that's wrong. Let me try again xD
1) (2tanx)/(sec^2x)

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Gasp!! I think I got it :) \[\frac{ 2tanx }{ 1+\tan^2x }\] \[\frac{ 2*\frac{ sinx }{ cosx } }{ \frac{ 1 }{ \cos^2x } }\] \[2*\frac{ sinx }{ cosx }*\frac{ \cos^2x }{ 1 }\] = simplify and cross out a few things 2sinxcosx
@ganeshie8 is this correct? :)
then you'd turn 2*sin(x)*cos(x) into sin(2x)
make sure to keep the right hand side the same the whole time
after you reach 2sinxcosx use the identity sin2x = 2sinxcosx and you'll have sin2x = sin 2x
wait... is there a typo somewhere? you had \[\frac{2tanx}{x+\tan^2x}\] in your original question and then all of a sudden I see \[\frac{2tanx}{1+\tan^2x}\] I am crossing fingers here because if that x+tan^2x turns out to be a typo then everything will work out
Right right, I just wanted to make sure the process was correct :)
um..the second one hahaa sorry yeah typo.
\[\frac{ 2tanx }{ 1+\tan^2x } = \sin2x \] \[\frac{ 2tanx }{ \sec^2x } =\sin 2x \] \[\frac{ 2\frac{sinx}{cosx} }{ \frac{1}{\cos^2x}} = sin2x \] \[2\frac{sinx}{cosx} \times \frac{\cos^2x}{1} = \sin 2x \] the cos x cancels out and 2/1 = 2 \[2sinxcosx = 2sinx \] using identity 2sinxcosx = sin2x \[\sin2x = \sin2x \]
your attempt is correct...just have to remember that there are A LOT of trig. identities.
Right...So I had it all right :) I just didn't do the last step to make it into sin2x.
thanks so much!!
very great work @Babynini
thank you :)

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