Problem:
If tanα and tanβ are the roots of x2−px+q=0, and cotα and cotβ are the roots of x2−rx+s=0, then rs is necessarily
Answer Choices:
A. pq
B. pq1
C. q2p
D. p2q
E. qp
Solution:
By the relationship between the roots and the coefficients of a quadratic equation, it follows that p=tanα+tanβ,q=tanαtanβ,r=cotα+ cotβ, and s=cotαcotβ. Since cotα+cotβ=tanα1+tanβ1=tanαtanβtanα+tanβ, and cotαcotβ=tanαtanβ1, the equalities r=qp and s=q1 follow. Thus rs=q2p.
Alternate solution. The roots are reciprocals, and in general, if x2−px+q=0 has roots m and n, then qx2−px+1=0 has roots m1 and n1. Dividing the last equation by q shows that x2−qpx+q1=0 also has roots m1 and n1. So r=qp,s=q1, and rs=q2p