Solution
= Solution
Since $\tan z=\sin z/\cos z$, its singularities occur at the simple zeros of the denominator,
$$
a_k=\frac{(2k+1)\pi}{2},
\qquad k\in\mathbb Z.
$$
At each one, $\sin a_k\ne0$ and $(\cos z)'|_{a_k}=-\sin a_k\ne0$, so all are simple poles. There are no other finite singularities:
$$
\boxed{\tan z\text{ is meromorphic with simple poles at }a_k}.
$$