What is the Cartesian form of $r^2-3r = 2theta+cot(theta)-tan(theta)$ ?

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Answer 1 · 2016-05-04T08:58:58

$x^2+y^2=2tan^(-1)(y/x)+x/y-y/x$

If $(r,theta)$ is in polar form and $(x,y)$ in Cartesian form the relation between them is as follows:

$x=rcostheta$ , $y=rsintheta$ , $r^2=x^2+y^2$ and $tantheta=y/x$ or and $cottheta=x/y$

Hence, $r^2-3r=2theta+cottheta-tantheta$ can be written as

$x^2+y^2=2tan^(-1)(y/x)+x/y-y/x$

What is the Cartesian form of r^2-3r = 2theta+cot(theta)-tan(theta) r^2-3r = 2theta+cot(theta)-tan(theta) ?