How do you convert #r = sec(θ + π/4)# into cartesian form? Precalculus Polar Coordinates Converting Equations from Polar to Rectangular 1 Answer Shwetank Mauria Jun 29, 2016 #x-y=sqrt2# Explanation: A polar coordinate #(r,theta)# in cartesian coordinates is #(x,y)# where #x=rcostheta# and #y=rsintheta#. It is apparent that #r^2=x^2+y^2#. Now #r=sec(theta+pi/4)=1/cos(theta+pi/4)=1/(costhetacos(pi/4)-sinthetasin(pi/4))=sqrt2/(costheta-sintheta)# (as #sin(pi/4)=cos(pi/4)=1/sqrt2# Hence, #rcostheta-rsintheta=sqrt2# or #x-y=sqrt2# Answer link Related questions What is the polar equation of a horizontal line? What is the polar equation for #x^2+y^2=9#? How do I graph a polar equation? How do I find the polar equation for #y = 5#? What is a polar equation? How do I find the polar equation for #x^2+y^2=7y#? How do I convert the polar equation #r=10# to its Cartesian equivalent? How do I convert the polar equation #r=10 sin theta# to its Cartesian equivalent? How do you convert polar equations to rectangular equations? How do you convert #r=6cosθ# into a cartesian equation? See all questions in Converting Equations from Polar to Rectangular Impact of this question 2823 views around the world You can reuse this answer Creative Commons License