We will turn the left side into the right side.
[tex] \dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x[/tex]
Use the identity:
[tex] \cos 2x = \cos^2 x - \sin^2 x [/tex]
[tex] \dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x[/tex]
[tex] \dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x[/tex]
Now use the identity
[tex] \sin^2 x + \cos^2 x = 1 [/tex] solved for sin^2 x and for cos^2 x.
[tex] \dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x [/tex]
[tex] \dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x [/tex]
[tex] \dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x[/tex]
[tex] \cot^2 x = \cot^2 x [/tex]
