# 1983 AHSME Problems/Problem 20

## Problem 20

If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$, and $\cot{\alpha}$ and $\cot{\beta}$ are the roots of $x^2 - rx + s = 0$, then $rs$ is necessarily

$\textbf{(A)} \ pq \qquad \textbf{(B)} \ \frac{1}{pq} \qquad \textbf{(C)} \ \frac{p}{q^2} \qquad \textbf{(D)}\ \frac{q}{p^2}\qquad \textbf{(E)}\ \frac{p}{q}$

## Solution

By Vieta's Formulae, we have $\tan(\alpha)\tan(\beta)=q$ and $\cot(\alpha)\cot(\beta)=s$. Recalling that $\cot\theta=\frac{1}{\tan\theta}$, we have $\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s$.

Also by Vieta's Formulae, we have $\tan(\alpha)+\tan(\beta)=p$ and $\cot(\alpha)+\cot(\beta)=r$, and again using $\cot\theta=\frac{1}{\tan\theta}$, we have $\tan(\alpha)+\tan(\beta)=r(\tan(\alpha)\tan(\beta))$. Using $\tan(\alpha)\tan(\beta)=q$ and $\tan(\alpha)+\tan(\beta)=p$, we therefore deduce that $r=\frac{p}{q}$, which yields $rs = \frac{1}{q}\cdot\frac{p}{q}=\frac{p}{q^2}$.

Thus, the answer is $\boxed{\textbf{(C)} \ \frac{p}{q^2}}$.