Chapter 6 - Analytic Trigonometry - Section 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 519: 83

$\frac{24}{7}$

1. Let $cos^{-1}(-\frac{3}{5})=t$ (t in quadrant II), we have $cos(t)=-\frac{3}{5}$ and $tan(t)=-\frac{4}{3}$ 2. $tan(2cos^{-1}(-\frac{3}{5}))=tan(2t)=\frac{2tan(t)}{1-tan^2(t)}=\frac{2(-\frac{4}{3})}{1-(-\frac{4}{3})^2}=\frac{24}{7}$