The map $\tan : ((-\frac \pi 2, + \frac \pi 2),d) \to (\mathbb R,\rho)$, where $d$ is the usual metric $d(x,y) = |x-y|$, is an isomorphism of metric spaces : it is bijective and $\rho(\tan x, \tan y) = d(x,y)$. So your question is equivalent to : "What is the completion of $(-\frac \pi 2, + \frac \pi 2)$ equipped with the usual distance ?"
Well, since $(-\frac \pi 2, + \frac \pi 2)$ is a subset (can I say a sub-metric space ?) of $\mathbb R$, which we know is complete, its completion is its closure in $\mathbb R$, which is $[-\frac \pi 2, + \frac \pi 2]$ : We only have to add two new points, corresponding to sequences converging to $\pm \frac \pi 2$, or in the original context, sequences diverging to $\pm \infty$.
Of course you can try to show this directly, but you will only hide all of this and make it harder.
Here is what you would have to prove :
- If a sequence $(y_n)$ diverges to $+ \infty$, then it is equivalent to your sequence $(x_n = n)$ : $\rho(x_n,y_n) = \arctan(n)- \arctan(y_n)$ and since both sequence diverge to $+ \infty$, $\arctan(n)$ and $\arctan(y_n)$ both converge to $\pi/2$, thus $\rho(x_n,y_n)$ converges to $0$.
- If a sequence $(y_n)$ diverges to $- \infty$, then it is equivalent to the sequence $(x_n = -n)$
- If a sequence $(y_n)$ converges to $a \in \mathbb R$, then it is equivalent to the constant sequence $(x_n = a)$. This results from the continuity of $\arctan$
- If a sequence is in none of the above three cases, it is not a Cauchy sequence. This is a bit painful and needs to use the fact that $\mathbb R$ is complete