How do you convert $r = sec (θ - \frac{π}{6})$ $r=sec(theta - pi/6)$ into cartesian form?

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Answer 1 · 2016-11-23T13:58:12

$\sqrt{3} x + y = 2$ $sqrt3x+y=2$

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

$x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$

Hence, $r = sec (θ - \frac{π}{6}) = \frac{1}{cos (θ - \frac{π}{6})}$ $r=sec(theta-pi/6)=1/cos(theta-pi/6)$

or $r = \frac{1}{cos θ cos (\frac{π}{6}) + sin θ sin (\frac{π}{6})}$ $r=1/(costhetacos(pi/6)+sinthetasin(pi/6))$

or $r (cos θ cos (\frac{π}{6}) + sin θ sin (\frac{π}{6})) = 1$ $r(costhetacos(pi/6)+sinthetasin(pi/6))=1$

or $r cos θ \times \frac{\sqrt{3}}{2} + r sin θ \times \frac{1}{2} = 1$ $rcosthetaxxsqrt3/2+rsinthetaxx1/2=1$

or $\sqrt{3} x + y = 2$ $sqrt3x+y=2$

How do you convert r=sec(theta - pi/6)r=sec(θ−π6)r=sec(theta - pi/6) into cartesian form?