Answer
$r = \frac{{2\cos \theta }}{{\cos \theta - \sin \theta }}$ as an equation in rectangular coordinates:
${x^2} + {y^2} = \frac{{4{x^2}}}{{{{\left( {x - y} \right)}^2}}}$
Work Step by Step
Since $r = \frac{{2\cos \theta }}{{\cos \theta - \sin \theta }}$ we have
${r^2} = \frac{{4{{\cos }^2}\theta }}{{{{\cos }^2}\theta - 2\cos \theta \sin \theta + {{\sin }^2}\theta }}$
Since $x = r\cos \theta $ and $y = r\sin \theta $, we get ${x^2} + {y^2} = {r^2}$.
$\cos \theta = \frac{x}{{\sqrt {{x^2} + {y^2}} }}$, ${\ \ \ }$ $\sin \theta = \frac{y}{{\sqrt {{x^2} + {y^2}} }}$.
So,
${x^2} + {y^2} = \frac{{4{x^2}/\left( {{x^2} + {y^2}} \right)}}{{{x^2}/\left( {{x^2} + {y^2}} \right) - 2xy/\left( {{x^2} + {y^2}} \right) + {y^2}/\left( {{x^2} + {y^2}} \right)}}$
${x^2} + {y^2} = \frac{{4{x^2}}}{{{x^2} - 2xy + {y^2}}}$
${x^2} + {y^2} = \frac{{4{x^2}}}{{{{\left( {x - y} \right)}^2}}}$
