Convert the polar equation to rectangular coordinates.

Solved it 

Initial Question:

\(sec(\theta )=2\)

Convert it to cosine

\(cos(\theta)=\frac{1}{2}\)

Which means

\(\theta = \pm \frac{\pi}{3}\)

\(tan(\theta)=\pm\frac{\sqrt{3}}{1}=\pm\sqrt{3}\)

\(\frac{y}{x}=\pm\sqrt{3}\)

Final answer is.

\(y=\pm\sqrt{3}x\)

sec(θ)=2

r/x  = 2

sqrt [ x^2 + y^2] / x  = 2

sqrt [ x^2 + y^2 ]  = 2x     square both sides

x^2 + y^2  = 4x^2

y^2  = 3x^2         take the square root of both sides

y =  ±√(3)x