8.49 If $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$ are the roots of the equation $x^{2}-p x+q=0$, and $\operatorname{ctg} \alpha$ and $\operatorname{ctg} \beta$ are the roots of the equation $x^{2}-r x+s=0$, then $r s$ must be
(A) $p q$.
(B) $\frac{1}{p q}$.
(C) $\frac{p}{q^{2}}$.
(D) $\frac{q}{p^{2}}$.
(E) $\frac{p}{q}$.
(34th American High School Mathematics Examination, 1983)
［Solution 1］By Vieta's formulas, we have
$$
p=\operatorname{tg} \alpha+\operatorname{tg} \beta, q=\operatorname{tg} \alpha \operatorname{tg} \beta \text {, and } r=\operatorname{ctg} \alpha+\operatorname{ctg} \beta, s=\operatorname{ctg} \alpha \operatorname{ctg} \beta \text {. }
$$

Therefore, $\operatorname{ctg} \alpha+\operatorname{ctg} \beta=\frac{1}{\operatorname{tg} \alpha}+\frac{1}{\operatorname{tg} \beta}=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{\operatorname{tg} \alpha \operatorname{tg} \beta}$, and
$$
\operatorname{ctg} \alpha \operatorname{ctg} \beta=\frac{1}{\operatorname{tg} \alpha \operatorname{tg} \beta} .
$$

Thus, $r=\frac{p}{q}, s=\frac{1}{q}, \quad$ which means $\quad r s=\frac{p}{q^{2}}$.
Therefore, the answer is $(C)$.
［Solution 2］Since the roots of these two equations are reciprocals of each other, generally speaking, if $x^{2}-$ $p x+q=0$ has roots $m$ and $n$, then $q x^{2}-p x+1=0$ will have roots $\frac{1}{m}$ and $\frac{1}{n}$.

Dividing both sides of $q x^{2}-p x+1=0$ by $q$, we get the equation $x^{2}-\frac{p}{q} x+\frac{1}{q}=0$, and $\frac{1}{m}$ and $\frac{1}{n}$ are the roots of this equation.

So $r=\frac{p}{q}, s=\frac{1}{q}$, and $r s=\frac{p}{q^{2}}$. Therefore, the answer is $(C)$.