Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 497: 42

$-\frac{\sqrt {15}}{7}$

Step 1. Using the given figures, we have $a=-\sqrt {5-4}=-1, b=-\sqrt {1-\frac{1}{16}}=-\frac{\sqrt {15}}{4}$ Step 2. $h(2\alpha)=tan(2\alpha)=\frac{2tan\alpha}{1-tan^2\alpha}=\frac{2(\sqrt {15})}{1-(\sqrt {15})^2}=-\frac{\sqrt {15}}{7}$