The arctangent of a standard Cauchy random variable $Z\sim\text{Cauchy}(0,1)$ is uniformly distributed in $[-\frac{\pi}{2},\frac{\pi}{2}]$. The proof is straightforward:
$$P(\arctan(Z)\leq t)=P(Z\leq \tan t)=\frac{t}{\pi}+\frac{1}{2}.$$
Now, standard Cauchy random variable is the ratio between two i.i.d. standard normal random variables, that is, if $X\sim\mathcal{N}(0,1)$ and $Y\sim\mathcal{N}(0,1)$, then $X/Y\sim\text{Cauchy}(0,1)$. The arctangent of $X/Y$ should thus also be uniformly distributed in $[-\frac{\pi}{2},\frac{\pi}{2}]$. However, when I evaluate the CDF explicitly, I obtain:
$$\begin{align} \tag{1}P(\arctan(X/Y)\leq t)&=P(X/Y\leq\tan t)\\ \tag{2}&=P(X\leq Y\tan t)\\ \tag{3}&=P(X-Y\tan t\leq 0)\\ \tag{4}&=\frac{1}{2}, \end{align}$$
where (4) is because $X-Y\tan t\sim\mathcal{N}(0,1+\tan^2t)$ by the additivity of i.i.d. normal random variables. Clearly I made a mistake somewhere, however, I am not sure where. Can anyone help?