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#.
