Check the differentiablity of $\theta :\mathbb R^2 \setminus\{(0,0)\}\to \mathbb R$

For , $(x,y)\in \mathbb R^2$ with $(x,y)\not =(0,0)$ , let $\theta=\theta(x,y)$ be the unique real number such that $-\pi<\theta \le\pi$ and $(x,y)=(r\cos \theta , r\sin \theta)$ , where $r=\sqrt{x^2+y^2}$. Then the resulting function $\theta :\mathbb R^2 \setminus\{(0,0)\}\to \mathbb R$ is

(A) differentiable.

(B) continuous but not differentiable.

(C) bounded but not continuous.

(D) neither bounded nor continuous.

Here , $\theta=\arctan\left(\frac{y}{x}\right)$. So it is differentiable in $\mathbb R^2 \setminus \{(0,0)\}$. So option (A) is correct.

Am I correct ?

Hint:

Trace the unit circle multiple times around and keep track of the value of $\theta$. How does it change when you're around the point $-1$? (Remember, it's the unique value of the angle between $-\pi$ and $\pi$.)