I know that the singular points of $\tan(z)$ occur for $\cos(z) = 0$, but I don't know the type. Typically, I determine the singularities of a function and their type with its Laurent series, which for tangent is $$ \tan(z) = z + \frac{z^3}{3} + \frac{2z^5}{15} + \cdots \quad\text{for } |z| < \frac{\pi}{2}. $$
Does this series expansion of $\tan(z)$ imply that the singularities are removable?
The zeroes of $g(z)=\cos z$ are simple zeroes, since the derivative $g'(z)=-\sin z$ does not vanish at those points: indeed, $\cos z = \frac{e^{iz}+e^{-iz}}{2}=0$ iff $e^{2 i z} =-1=e^{i \pi}$ so iff $2iz=i\pi +2\pi ik$, i.e. $z=\pi/2+k\pi$, $k \in \mathbb{Z}$, and in all those points $g'(z)=\pm1 \neq 0$; hence your singularities are simple poles, since they are zeroes of order $0$ (i.e. they are not zeroes nor poles) of the numerator and simple zeroes of the denominator.