What is the Cartesian form of #r^2-2r = -theta+tan(theta)- cos^2(theta) #? Trigonometry The Polar System Converting Between Systems 1 Answer Sonnhard · Stefan V. Jul 17, 2018 #x^2+y^2-2sqrt(x^2+y^2)=-arctan(y/x)+y/x-x^2/(x^2+y^2)# Explanation: We are Using #x=r cos(theta)# #y=r sin(theta)# #theta=arctan(y/x)# #r=sqrt(x^2+y^2)# so we get #x^2+y^2-2sqrt(x^2+y^2)=-arctan(y/x)+y/x-x^2/(x^2+y^2)# Answer link Related questions How do you convert rectangular coordinates to polar coordinates? When is it easier to use the polar form of an equation or a rectangular form of an equation? How do you write #r = 4 \cos \theta # into rectangular form? What is the rectangular form of #r = 3 \csc \theta #? What is the polar form of # x^2 + y^2 = 2x#? How do you convert #r \sin^2 \theta =3 \cos \theta# into rectangular form? How do you convert from 300 degrees to radians? How do you convert the polar equation #10 sin(θ)# to the rectangular form? How do you convert the rectangular equation to polar form x=4? How do you find the cartesian graph of #r cos(θ) = 9#? See all questions in Converting Between Systems Impact of this question 1727 views around the world You can reuse this answer Creative Commons License