Question

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

Answer 1

Just a nuance in the answer:
`-2 +x^2+ y^2 = - (pi + arctan ( y/x)) +(xy)/( x^2 + y^2 )`,
when `theta in Q_3, Q_2`.

Explanation:

`arctan(y/x) in ( -pi/2, pi/2 )`, and so, constrained not to give

`theta in Q_2 and Q_3` .

Examples:

If `theta = 3/4pi, arctan( y/x) = arctan(-1) = - pi/4`

`and pi + ( - pi/4 ) = 3/4pi`.

If `theta = 5/4pi, arctan ( y/x ) = arctan ( 1 ) = pi/4`,

`and pi + pi/4 = 5/4pi`.

I removed the arctan operator, for tan, and got the overall inverse

y/x = tan(2 -(x^2+y^2)+(xy)/(x^2+y^2)).

The graph is bewildering.

graph{y/x-tan(2 -(x^2+y^2)+(xy)/(x^2+y^2))=0}