Residue of $\sin(\tan(z))$ at $z=\frac{\pi}{2}$ In finding this residue, I took a Taylor expansion of $\sin(x)$ and substituted in $\tan(z)$ and then attempted to compute the residues of the odd powers of $\tan(z)$ at $z=\frac{\pi}{2}$. Using Wolfram Alpha, it seemed to be that: 
\begin{equation}
\operatorname{Res}(\tan^{2n-1}(z), \frac{\pi}{2}) = (-1)^n
\end{equation} 
Giving the residue of $\sin(\tan(z))$ as $-\sinh(1)$.
Is this correct?
Also, I was unable to show $\operatorname{Res}(\tan^{2n-1}(z), \frac{\pi}{2}) = (-1)^n$.
Is this true and if so any hints on how to show it?
 A: Using $\sin\tan z=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}\left(\tan z\right)^{2n+1}$ is a good idea. Then
$$ \operatorname*{Res}_{z=\pi/2}\left(\tan z\right)^{2n+1}= -\operatorname*{Res}_{z=0}\left(\cot z\right)^{2n+1}=-\frac{1}{2\pi i}\oint_{\|z\|=\varepsilon}\left(\cot z\right)^{2n+1}\,dz$$
by Cauchy's integral formula. The substitution $z=\arctan u$ allows to write the RHS as
$$ -\frac{1}{2\pi i}\oint_{\|u\|=\delta}\frac{du}{(u^2+1)u^{2n+1}}=-\operatorname*{Res}_{u=0}\frac{1}{u^{2n+1}}\left(1-u^2+u^4-u^6+\ldots\right) $$
hence
$$ \operatorname*{Res}_{z=\pi/2}\left(\tan z\right)^{2n+1}= (-1)^{n+1}, $$
$$\operatorname*{Res}_{z=\pi/2}\sin\tan z=-\sum_{n\geq 0}\frac{1}{(2n+1)!}=-\sinh(1).$$
A: As a complement to Jack's answer, we can write for any $\epsilon \in (0,\pi/2)$
$$\begin{align}
\oint_{|z-\pi/2|=\epsilon}\sin(\tan(z))\,dz&\overbrace{=}^{z\mapsto z+\pi/2}-\oint_{|z|=\epsilon}\sin(\cot(z))\,dz\\\\
&\overbrace{=}^{z\mapsto \arctan(z)}-\oint_{|z|=\tan(\epsilon)}\frac{\sin(1/z)}{1+z^2}\,dz\\\\
&=-\oint_{|z|=\tan(\epsilon)}\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!z^{2n+1}}\,\sum_{m=0}^\infty (-1)^mz^{2m}\,dz\\\\
&=-\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!}\,\sum_{m=0}^\infty (-1)^m\oint_{|z|=\tan(\epsilon)}z^{2m-2n-1}\,dz\\\\
&=-\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!}\,\sum_{m=0}^\infty (-1)^m 2\pi i \delta_{mn}\\\\
&=-2\pi i \sum_{n=0}\frac{1}{(2n+1)!}\\\\
&=2\pi i (-\sinh(1))
\end{align}$$
as expected!
