How do you convert $r=sec(theta - pi/6)$ into cartesian form?

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Answer 1 · 2016-11-23T13:58:12

$sqrt3x+y=2$

Polar coordinates $(r,theta)$ and Cartesian oordinates $(x,y)$ are related as

$x=rcostheta$ and $y=rsintheta$

Hence, $r=sec(theta-pi/6)=1/cos(theta-pi/6)$

or $r=1/(costhetacos(pi/6)+sinthetasin(pi/6))$

or $r(costhetacos(pi/6)+sinthetasin(pi/6))=1$

or $rcosthetaxxsqrt3/2+rsinthetaxx1/2=1$

or $sqrt3x+y=2$

How do you convert r=sec(theta - pi/6)r=sec(theta - pi/6) into cartesian form?