How do you convert #r = sec(θ + π/4)# into cartesian form?

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#

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