Chapter 6 - Analytic Trigonometry - Section 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 519: 61

See below.

$RHS=\frac{1-tan^2\frac{\theta}{2}}{1+tan^2\frac{\theta}{2}} =\frac{1-\frac{1-cos\theta}{1+cos\theta}}{1+\frac{1-cos\theta}{1+cos\theta}} =\frac{1+cos\theta-(1-cos\theta)}{1+cos\theta+(1-cos\theta)} =\frac{2cos\theta}{2}=cos\theta=LHS$