Find $\displaystyle {\tan{{\left(\alpha-\beta\right)}}}$ exactly, given $\displaystyle {\sec{\alpha}}=\frac{{5}}{{3}}$ and $\displaystyle {\sin{\beta}}=\frac{{6}}{{7}}$ where $\displaystyle \alpha$ terminates in quadrant IV and $\displaystyle \beta$ terminates in quadrant II. Be sure to reduce your final result completely.