Question

What is the Cartesian form of `r = -sin^2theta-thetacsc^2theta`?

Answer 1

See answer in the explanation.

Explanation:

Using

`0 <= r = sqrt ( x^2 + y^2 ), r ( cos theta, sin theta ) = ( x, y )`.

`theta = arctan (y/x ), theta in Q_1` or `Q_4` and

`theta = pi + arctan (y/x ), theta in Q_2` or `Q_3`

`r = - sin^2theta -theta csc^2theta` converts to

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

where

`phi = arctan (y/x ), theta in Q_1` or `Q_4` and

`= pi + arctan (y/x ), theta in Q_2` or `Q_3`

See graph.
graph{y^2(x^2+y^2)^1.5+y^4 + (x^2+y^2)^2 arctan (y/x )=0}
Graph for the wholesome inverse.
graph{tan((y^2(x^2+y^2)^1.5+y^4)/(x^2+y^2))-1=0}