Question

What is the Cartesian form of `rtheta+r = 2costheta-cottheta*sectheta`?

Answer 1

`"Cartesian form of"`
`rtheta+r=2costheta-cotthetaxxsectheta`
`" is given by "`
`(sqrt(x^2+y^2))(1+tan^-1(y/x))-(sqrt(x^2+y^2))/y`

Explanation:

`rtheta+r=2costheta-cotthetaxxsectheta`

`x=rcostheta`

`y=rsintheta`

`r=sqrt(x^2+y^2)`

`theta=tan^-1(y/x)`

`tantheta=y/x`

`costheta=x/r=x/sqrt(x^2+y^2)`

`cottheta=1/tantheta=1/(y/x)=x/y`
`sectheta=1/costheta=1/(x/sqrt(x^2+y^2))=(sqrt(x^2+y^2))/x`

Thus,

`rtheta+r=2costheta-cotthetaxxsectheta " becomes"`

`(sqrt(x^2+y^2))(tan^-1(y/x))+(sqrt(x^2+y^2))=2xxx/sqrt(x^2+y^2)-x/yxx(sqrt(x^2+y^2))/x`
#

Simplifying

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