Answer
$r^{2}-\sec 2\theta=0$
Work Step by Step
$ x=r\cos\theta$ and $ y=r\sin\theta$
$r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$.
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$x^{2}-y^{2}=1 $
Substitute $ r\cos\theta$ for $x, r\sin\theta$ for $y.$
$(r\cos\theta)^{2}-(r\sin\theta)^{2}=1$
$r^{2}(\cos^{2}\theta-\sin^{2}\theta)=1$
... recognize a double angle identity...
$r^{2}\cos 2\theta=1$
$r^{2}=\displaystyle \frac{1}{\cos 2\theta}$
$ r^{2}=\sec 2\theta$
$r^{2}-\sec 2\theta=0$
