Chapter 6 - Analytic Trigonometry - Section 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 519: 89

$\frac{25}{7}$

1. Let $tan^{-1}(\frac{3}{4})=t$, we have $tan(t)=\frac{3}{4}$ and $cos(t)=\frac{4}{5}$ 2. $sec(2tan^{-1}(\frac{3}{4}))=sec(2t)=\frac{1}{cos(2t)}=\frac{1}{2cos^2t-1}=\frac{1}{2(\frac{4}{5})^2-1}=\frac{25}{7}$