Chapter 6 - Section 6.3 - Trigonometric Equations Involving Multiple Angles - 6.3 Problem Set - Page 342: 61

$$\frac{\sin x}{1+\cos x} =\frac{ 1-\cos x }{\sin x}$$

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