Residue of $\sec(1/z)$ at $z=0$
It makes no sense to talk about the residue of $\sec\left(\frac1z\right)$ at $0$. That concept is only defined for isolated singularities, and $0$ is not one such singularity, since $\sec\left(\frac1z\right)$ has singularities arbitrarily close to $0$; just consider the numbers of the form $\left(\frac\pi2+k\pi\right)^{-1}$, with $k\in\Bbb Z$.