Question on proving an identity

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The discussion revolves around proving the identity 2csc(2x) = csc^2(x)tan(x). The user initially simplifies the left side to 2(2sin(x)cos(x)) but struggles to connect it to the right side. A response clarifies that 2(2sin(x)cos(x)) can be rewritten as (1/sin^2(x)) multiplied by sin(x)cos(x), leading to the desired identity. The user expresses gratitude after gaining a clearer understanding of the transformation. The exchange highlights the importance of manipulating trigonometric identities to prove equations.
Lazz72
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So I'm given a problem in which I have to prove an identity. It goes:
2csc2x=csc^2xtanx

I did the problem myself and could only get to 2csc2x=2\(sin2x)= 2\(2sinxcosx). I had no idea how to get further with the problem so I looked at the answer in the back of my pre-calculus book. It said that:

2\(2sinxcosx) = 1\(sin^2x) multiplied by sinx\cosx = csc^2xtanx

I don't understand the part that says 2\(2sinxcosx)= 1\(sin^2x) multiplied by sinx/cosx. Could someone please explain to me how 2\(2sinxcosx) equals that?
 
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Hi Lazz72, welcome to PF.
2csc(2x) = 2/sin(2x) = 2/2six*cosx = 1/sinx*cos* = sinx/sinx*sinx*cosx = tanx/sin^2(x)
 
Ahh okay I understand it now, thanks!
 

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