Answer
$-\frac{1}{7}$
Work Step by Step
1. Let $cos^{-1}(\frac{3}{5})=u$ (quadrant I), we have $cos(u)=\frac{3}{5}, tan(u)=\frac{4}{3}$
2. $tan(\frac{\pi}{4}-u)=\frac{tan\frac{\pi}{4}-tan(u)}{1+tan\frac{\pi}{4}tan(u)}
=\frac{1-\frac{4}{3}}{1+(\frac{4}{3})}=-\frac{1}{7}$
