In order to transform the polar coordinates #(r_P,theta_P)# of a point #P# to Cartesian coordinates #(x_P,y_P)# you must solve the system of equations :
#x_P^2 + y_P^2 = r_P^2#
#y_P = r_P sin(theta_P)#
#x_P = r_P cos(theta_P)#
Now, let #P# be a point on the graph of of the function #r^2 = 2theta+ cot(theta)-tan(theta)#. The point #P# is defined by the equation
#r_P^2 = 2theta_P + cot(theta_P) - tan(theta_P)#.
We know from the first equation of the system that #color(red)(r_P^2) = x_P^2 + y_P^2#.
From the second and third, you can get #color(red)(theta_P)# to be equal to #arcsin(y_P/sqrt(x_P^2 + y_P^2))# and #arccos(x_P/sqrt(x_P^2+y^P2))#,
having already substituted #r_P#. To simplify things, let
#color(blue)(alpha_P) = y_P/sqrt(x_P^2 + y_P^2)# and #color(blue)(beta_P)# = #x_P/sqrt(x_P^2 + y_P^2)#.
Now, substitute #color(red)(theta_P)# and #color(red)(r_P)# into the formula that defines #P# :
#r_P^2 = 2theta_P + cot(theta_P) - tan(theta_P)#
#x_P^2 + y_P^2 = 2arcsin(color(blue)(alpha_P)) + (cos(color(red)(arccos)(color(blue)(beta_P))))/(sin(color(red)(arcsin)(color(blue)(alpha_P)))) + (sin(color(red)(arcsin)(color(blue)(alpha_P))))/cos(color(red)(arccos)(color(blue)(beta_P)))#
We have substitued #theta_P# in such a way that it is most convienient. However, there was no convinient way to write #2theta_P#.
#x_P^2 + y_P^2 = 2arcsin(color(blue)(alpha_P)) + (color(blue)(beta_P))/(color(blue)(alpha_P)) + (color(blue)(alpha_P))/(color(blue)(beta_P))#
#color(blue)(beta_P)/color(blue)(alpha_P)# and it's reciprocal cancel each other pretty neatly :
#x_P^2 + y_P^2 = 2arcsin(color(blue)(y_P/sqrt(x_P^2 + y_P^2))) + color(blue)(x_P)/color(blue)(y_P) + color(blue)(y_P)/color(blue)(x_P)#.
And finally, the equation that defines #P# in the Cartesian coordinate system is
#x_P^2 + y_P^2 = 2arcsin(y_P/sqrt(x_P^2 + y_P^2)) + (x_P)/(y_P) + (y_P)/(x_P)#.
Since #P# is just a random point on the graph, we can generelise the answer to be
#color(blue)(x^2 + y^2 = 2arcsin(color(blue)(y/sqrt(x^2 + y^2))) + color(blue)(x)/color(blue)(y) + color(blue)(y)/color(blue)(x)#
