How do you convert $r= tan θ$ to rectangular form?

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Answer 1 · 2016-08-26T02:21:06

$y=+-x^2/sqrt(1-x^2), x in(-1, 1)$ . The graph is through the origin and is symmetrical about the origin. As $x to +-1, y to +-oo$

Use the conversion formula $r(cos theta, sin theta) = (x, y)$ .that givs

$r = sqrt(x^2+y^2), cos theta = x/r and sin theta = y/r$

Here, the conversion gives

$sqrt(x^2+y^2) =y/x$ or explicitly, after solving for y,

$y =+-x^2/sqrt(1-x^2)$ and for real $y, x in (-1. 1)$

Not to miss is my observation that the periodicity characteristic of

$tan theta, theta$ , with period $pi$ , for $theta in ( -oo, oo )$ in polar

form $r = tan theta$ is not explicit in the Cartesian form

$y =+-x^2/sqrt(1-x^2), x in (-1. 1)$

How do you convert r= tan θr= tan θ to rectangular form?