Question

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

Answer 1

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

Explanation:

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`