= Solution
Put
$$
z_k=\frac1{a_k}=\frac{2}{(2k+1)\pi},
\qquad w=z-z_k.
$$
Near $a_k$, writing $\delta=1/z-a_k$ gives
$$
\tan(a_k+\delta)=-\cot\delta=-\frac1\delta+\frac\delta3+O(\delta^3).
$$
But
$$
\delta=\frac1z-\frac1{z_k}=-\frac{w}{z_k(z_k+w)},
$$
and hence, exactly,
$$
-\frac1\delta=\frac{z_k(z_k+w)}w=\frac{z_k^2}{w}+z_k.
$$
Since $\delta=O(w)$, the first two Laurent terms are
$$
\boxed{
\tan\frac1z
=\frac{z_k^2}{z-z_k}+z_k+O(z-z_k)}.
$$
Equivalently, substitute $z_k=2/((2k+1)\pi)$ in this expression.
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