What is the functional form of $r = e^{\tan(\theta)}$?

Question

$r = e^{\tan(\theta)}$ generates a functional curve. How can this be converted to cartesian form? Or, is it transcendental?

What I've got so far:
https://www.desmos.com/calculator/vbc3muenfo
The parametric form of the curve is $ ( e^{\tan(t)}\cos(t), e^{\tan(t)}\sin(t) )$. Both $x(t)$ and $y(t)$ are periodic with seemingly functional periods. Can a single-branched inverse resembling the inverse trigonometric functions be defined on some period?

UPDATE: Parametric form might not be necessary.

When converting polar equations to cartesian equations, $r = \sqrt{x^2+y^2}$ and $\tan(\theta)=\frac yx$ by the definition of the tangent function and the pythagorean theorem. Therefore, by substitution:

$r = e^{\tan(θ)} \Rightarrow \sqrt{x^2+y^2}=e^\frac yx$

How can this be converted to a function of the form, $y=f(x)$?
Visual: https://www.desmos.com/calculator/plngic6rdq

Answer 1

not sure if that right, but you have that (catersian and polar relations) $$x=r\cos\theta,y=r\sin\theta$$ so divide both to have $$\frac{y}{x}=\frac{r\sin\theta}{r\cos\theta}=\tan\theta$$ and $$x^2+y^2=(r\cos\theta)^2+(r\sin\theta)^2=r^2(\cos^2\theta+\sin^2\theta)=r^2$$ so $$r=e^{\tan\theta}\\ r^2=e^{2\tan\theta}\\ x^2+y^2=e^{\frac{2y}{x}}$$ or considering $r>0$ we can assume that $\sqrt{r^2}=r$ so $$\sqrt{x^2+y^2}=e^{\frac{y}{x}}$$

Answer 2

\begin{eqnarray} y&=&x\tan\theta\\ &=&x\ln(r)\\ &=&\frac{1}{2}x\ln\left(x^2+y^2\right) \end{eqnarray}