Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.3 Polar Coordinates - Exercises - Page 618: 20

$$ r= \sqrt{\sec \theta\csc\theta} $$

Given $$ xy=1$$ Since $ y=r\sin\theta,\ \ x= r\cos\theta,\ \ r^2=x^2+y^2 $, then \begin{align*} xy&=1\\ r^2\sin \theta\cos\theta&=1 \\ r^2&=\sec \theta\csc\theta \\ \end{align*} Hence $$ r= \sqrt{\sec \theta\csc\theta} $$