How to get correct angle when using inverse tangent for quadrant II or quadrant III?

Question

Whenever I calculate inverse tangent on the calculator it only gives the correct angle when the point (X, Y) is in the first and fourth quadrant. Whenever the point lies in the second or third quadrant the answer is wrong, how do I correct it, when it lies in one of those quadrants? I am surprised I was not able to find an answer on Google, maybe it is because I am wording the problem awkwardly, but I know there has to be an explanation for this. I am trying to convert from rectangular coordinates into polar coordinates and this problem keeps coming up consistently.

Answer 1

We knon that $\tan(x)$ is not a bijective function in all the domain, so to find the inverse, we have to restrict the domain to $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. This explain why we can't obtain angles as $\pi$ or $-\frac{3\pi}{4}$ that we find in third and second quadrant. To overcome this, we can add $\pi$ for tge angles in second quadrant. So: $$\alpha=\arctan(v)+\pi$$ where $v\leq0, v\in R$ and so $\arctan(v)\leq0$ For the angles in third quadrant, we have: $$\beta=\arctan(v)+\pi$$ where $v\geq 0$ and so $\arctan(v)\geq 0$.