If $\tan A=\frac{p}{q}$ for an acute angle A, express $\frac{p\sin A-q\cos A}{p\sin A+q\cos A}$ in terms of $p$ and $q$.

Divide numerator and denominator by $\cos A:\frac{p\tan A-q}{p\tan A+q}=\frac{p\cdot \frac{p}{q}-q}{p\cdot \frac{p}{q}+q}=\frac{(p^2-q^2)/q}{(p^2+q^2)/q}=\frac{p^2-q^2}{p^2+q^2}$. Answering $\frac{p-q}{p+q}$ comes from cancelling $\sin A$ against $\cos A$ as if they were equal; the last option reverses the order of subtraction in the numerator.