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

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Answer 1 · 2018-08-03T16:00:28

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#.

#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}