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#