If $\tan A=\frac{3}{4}$ for an acute angle A, find $\frac{\sec A+\tan A}{\sec A-\tan A}$.

${\sec }^{2}A=1+\frac{9}{16}=\frac{25}{16}$, so $\sec A=\frac{5}{4}$ and the ratio is $\frac{5/4+3/4}{5/4-3/4}=\frac{2}{1/2}=4$. Shortcut: because $(\sec A+\tan A)(\sec A-\tan A)=1$, the ratio equals $(\sec A+\tan A{)}^2=2^2=4$. Answering $\frac{1}{4}$ inverts the fraction; answering 2 stops at $\sec A+\tan A$ without squaring.