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

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Answer 1 · 2016-06-29T03:51:47

$x-y=sqrt2$

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

$x-y=sqrt2$

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