$\dfrac{1+\cos(2x)}{1-\cos(2x)} = \dfrac{1+\cos(2x)}{2} \cdot \dfrac{2}{1-\cos(2x)} = \dfrac{\cos^2{x}}{\sin^2{x}} = \cot^2{x}$
skeeter used the identities that $\displaystyle cos(2x)= 2 cos^2(x)- 1$ so that $\displaystyle 1+ cos(2x)= 2 cos^2(x)$ and $\displaystyle cos(2x)= 1- 2 sin^2(x)$ so that $\displaystyle 1- cos(2x)= 2sin^2(x)$.