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Condition (iii) is also sufficient. If $|a_n|$ did not tend to infinity, some $M>0$ would contain infinitely many terms with $|a_n|\le M$. Those terms form a bounded subsequence, and the Bolzano-Weierstrass theorem would give it a convergent subsequence, contrary to the hypothesis. Hence $|a_n|\to \infty$ and

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. It is enough to select infinitely many terms of the tangent sequence that stay uniformly bounded.

For each positive integer $k$, choose $n_k=\lfloor k\pi +1/2\rfloor$, using the floor function. Then $|n_k-k\pi |\le 1/2$. Moreover $n_{k+1}-n_k\ge \pi -1>0$, so these are strictly increasing indices. By the period and monotonicity of the tangent on $(-\pi /2,\pi /2)$,Thus $(\tan n_k)$ is a bounded subsequence. Applying the Bolzano-Weierstrass theorem to it gives a convergent subsequence, which is also a subsequence of the original sequence. Such a convergent subsequence therefore exists.