proof that left side = right side

First thing to notice is that the left side has double angles whereas the right hasn't. So, you can start by breaking the right side down to single angles.

Well I try and I couldn't solve it
I try both sides and I couldn't get the right answer

Could you post what you tried?

(sin2x)^2/1-cos2x=(2sinxcosx)(2sinxcosx)/1-(1-2(sinx)^2)

I try this side and on the half way I stopped

(sin2x)^2/1-cos2x=(2sinxcosx)(2sinxcosx)/1-(1-(sinx)^2)

Remember that: \cos(2x) = 1 - 2\sin^2(x)

2(sinx)^2 =2sin^2(x) yes I know

Okay, now expand the denominator. What do you get?

(2sinxcosx)(2sinxcosx)/1-1+2sin^2(x)
=(2sinxcosx)(2sinxcosx)/2sin^2(x)

Good. Now what can you cross out?

(2sinxcosx)(2sinxcosx)=2sin^2(x) the 2sin^2(x) with 2sin(x) ?

(4sinxcosx)/2sin^2(x) =2cos(x)/sin(x) that's what I get

(2sinxcosx)(2sinxcosx)=2sin^2(x) the 2sin^2(x) with 2sin(x) ?

I don't understand what you mean here...

\LARGE{\frac{(2\sin(x)\cos(x))(2\sin(x)\cos(x))}{2 \sin^2(x)} = \frac{(2\sin(x)\cos(x))(2\sin(x)\cos(x))}{2\sin(x) \sin(x)}}

(4sinxcosx)/2sin^2(x) =2cos(x)/sin(x) thats what I get

The bolded part is wrong.

What do u mean ?

I mean:

\Large{(2\sin(x)\cos(x))(2\sin(x)\cos(x))\ \not{=}\ 4\sin(x)\cos(x)}

sorry its 4sin^2(x)cos^2(x)

Okay, so now you have:

\LARGE{\frac{4\sin^2(x)\cos^2(x)}{2\sin^2(x)}}

What can you simplify?

4sin^2(x)cos^2(x)/2sin^2(x) then if I canceled the 4sin^2(x) with 2 sin^2(x) we end up with 2 cos^2(x) is this right ?

If you ended up with your proof, then sure! :D

Thank you :)

It wasn't that difficult, was it? ;)

no it wasn't
but I thought that we can not cancel because we have 4sin^2(x) cos^2(x) and on the denominator we have 2 sin^2(x) we miss the cos^2(x) that's why I was stuck with this problem