We have to prove this identity.
We will use the right side and rearrange it into the left side.
We can do it as:
[tex]\begin{gathered} 2\frac{\cos^4x}{\sin2x}-\cos^2x\cdot\csc2x-\frac{2\sin^2x\cdot\cos^2x}{\sin2x}+\sin^2x\cdot\csc2x \\ \\ (\cos^2x-\sin^2x)(\frac{2\cos^2x}{\sin2x}-\csc2x) \\ \\ (\cos2x)(\frac{2(\cos2x+\sin^2x)}{\sin2x}-\frac{1}{\sin2x}) \\ \\ (\cos2x)(2\frac{\cos2x}{\sin2x}+2\frac{\sin^2x}{\sin2x}-\frac{1}{\sin2x}) \\ \\ (\cos2x)(2\cot2x+\frac{2}{\sin2x}(\frac{1-\cos2x}{2})-\frac{1}{\sin2x}) \\ \\ (\cos2x)(2\cot2x+\frac{1-\cos2x}{\sin2x}-\frac{1}{\sin2x}) \\ \\ (\cos2x)(2\cot2x+\frac{1}{\sin2x}-\frac{\cos2x}{\sin2x}-\frac{1}{\sin2x}) \\ \\ (\cos2x)(2\cot2x-\frac{\cos2x}{\sin2x}) \\ \\ (\cos2x)(2\cot2x-\cot2x) \\ \\ (\cos2x)(\cot2x) \end{gathered}[/tex]
