Here is a polar equation $r = \tan(\theta) $. I converted it into the rectangular form. This would be
$$ \sqrt{x^2 + y^2} = \frac{y}{x} $$
When I plot both in Wolfram Alpha, I get different plots. Here is the polar plot and here is the rectangular plot. They are different. In polar plot, we see that $ \theta $ is restricted between $-\pi/2$ to $\pi/2$. Could this be reason for the different plots ? I don't know if Wolfram Alpha is correct or not ?
Part of the reason for the difference in the graphs is the that the polar plot shows only the plot for $-\frac\pi2 < \theta < \frac\pi2.$ If you plot the graph for $\frac\pi2 < \theta < \frac{3\pi}2$, it is the mirror image of the $-\frac\pi2 < \theta < \frac\pi2$ graph reflected across the $x$-axis.
Adding other values of $\theta$ will just repeat parts of these two graphs.
For the rectangular coordinates, the formula $\sqrt{x^2 + y^2}$ is equal to $r$ only when $r$ is non-negative. This leaves out all the parts of the polar graph where $r$ is negative, such as $-\frac\pi2 < \theta < 0$ and $\frac\pi2 < \theta < \pi.$ Those are the parts of the polar graph in the second and fourth quadrants. For example, all the parts of the polar graph in the second quadrant come from angle values in the fourth quadrant, which plot into the second quadrant because $\tan(\theta) < 0.$