How do you convert # r=-2 theta - tan theta # to Cartesian form? Trigonometry The Polar System Converting Between Systems 1 Answer Shiva Prakash M V Feb 18, 2018 #sqrt(x^2+y^2)=-(y/x+2tan^-1(y/x))# Explanation: #"Given,"r=-2theta-tantheta# #x=rcostheta, y=rsintheta# #tantheta=y/x# #r=sqrt(x^2+y^2), theta=tan^-1(y/x)# Thus, #r=-2theta-tantheta " becomes, "# # sqrt(x^2+y^2)=-2tan^-1(y/x)-y/x# Rearranging #sqrt(x^2+y^2)=-(y/x+2tan^-1(y/x))# 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 1769 views around the world You can reuse this answer Creative Commons License