If tanα and tanβ are the roots of x^2-px+q=0, and cotα and cotβ are the roots of x^2-rx+s=0, then rs is necessarily

1983 AMC 12 Problem 20

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20.

\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

pq

\frac1{pq}

\frac{p}{q^2}

\frac{q}{p^2}

\frac pq

The roots of the second quadratic are reciprocals of the roots of the first

Use Vieta’s formulas to express their sum and product in terms of

p,q

Let the first roots be

u=\tan\alpha

and

v=\tan\beta.

Then

u+v=p

and

uv=q.

The second roots are

\frac{1}{u}

and

\frac{1}{v},

\begin{aligned} r&=\frac1u+\frac1v=\frac pq,\\ s&=\frac1{uv}=\frac1q. \end{aligned}

Therefore

rs=\frac{p}{q^2}.

Therefore, the correct answer is C.