Chapter 6 - Section 6.3 - Trigonometric Equations Involving Multiple Angles - 6.3 Problem Set - Page 342: 62

$$\frac{\sin^2x}{(1-\cos x)^2} =\frac{1+\cos x}{ 1-\cos x } $$

Using $\cos ^2x+\sin ^2x=1$ \begin{align*} \frac{\sin^2x}{(1-\cos x)^2}&=\frac{1-\cos^2x}{(1-\cos x)^2}\\ &=\frac{(1-\cos x)(1+\cos x)}{(1-\cos x)^2}\\ &=\frac{1+\cos x}{ 1-\cos x } \end{align*} Then $$\frac{\sin^2x}{(1-\cos x)^2} =\frac{1+\cos x}{ 1-\cos x } $$