In order to transform the Polar Coordinates (r,theta) into (x,y), you must look at the triangle formed by origin, the point with coordinates (x,y) and the point (x,0).
This is a right angled triangle with hypotenuse equal to |r| and sides equal to |x| and |y|, respectively.
From Pythagora's Theorem and the definitions of sin x and cos x we have :
r^2 = x^2 + y^2
x = r*cos theta
y = r*sin theta
From these we can deduce theta to be equal to arccos(x/(x^2+y^2)) and arcsin(y/(x^2+y^2)).
Now, insert x^2 + y^2 and the formulas for theta into the original equation.
(x^2 + y^2)^2 + x^2 + y^2
= -arccos(x/(x^2+y^2)) - 1/cos^2 (arccos(x/(x^2 + y^2))) + 4/sin (arcsin(y/(x^2+y^2)))
=-arccos(x/(x^2 + y^2)) - (x^2 + y^2)^2/x^2 + (4(x^2+y^2))/y
So, in conclusion, the Cartesian form of r^2 + r = -theta - sec^2 theta + 4 csc theta
is
=-arccos(x/(x^2 + y^2)) - (x^2 + y^2)^2/x^2 + (4(x^2+y^2))/y
or
=-arcsin(y/(x^2 + y^2)) - (x^2 + y^2)^2/x^2 + (4(x^2+y^2))/y.
