Question

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

Answer 1

`x^2+y^2-2sqrt(x^2+y^2)=-arctan(y/x)+y/x-x^2/(x^2+y^2)`

Explanation:

We are Using

`x=r cos(theta)`

`y=r sin(theta)`
`theta=arctan(y/x)`
`r=sqrt(x^2+y^2)`

so we get

`x^2+y^2-2sqrt(x^2+y^2)=-arctan(y/x)+y/x-x^2/(x^2+y^2)`