Why is the reciprocal of the power series $1-\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots$ equal to $1+\frac{z^2}{6}+\frac{7z^4}{360}+\cdots$? \begin{align}
\frac1{\sin(z)}
&=\frac1z\frac{z}{\sin(z)}\\
&=\frac1z\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)^{-1}\\
&=\frac1z\left(1+\frac{z^2}{6}+\frac{7z^4}{360}+\frac{31z^6}{15120}+\cdots\right)\\
&=\frac1z+\frac{z}{6}+\frac{7z^3}{360}+\frac{31z^5}{15120}+\cdots
\end{align}
In particular, from lines 2 to 3, what is happening? (eg. why does reciprocating result in a change of signs?)
 A: They simply performed the division of $1$  by $1-\dfrac{z^2}{3!}+\dfrac{z^4}{5!}-\dfrac{z^6}{7!}+\dotsm$ by increasing powers.
For those who do not know division by increasing powers:
It is a long division in which the dividend and the divisor are ordered by increasing powers. One begins by dividing the lowest degree term of the dividend by the lowest degree term of the divisor, instead of the highest degree terms for Euclidean division. Contrary to the latter, division by increasing powers never stops. It is based on the following general result:

Given two polynomials $f,g \in K[X]$ $\:(g(0)\ne 0)$, for any $n\ge 0$, there  exists polynomials $q_n, r_n$ such that
  $$f(X)=q_n(X)g(X)+X^n r_n(X)$$
  and these polynomials are unique (for a given $n$)

This result can be extended to formal power series (it is this extension which is used here).
A detailed example in my answer to  this question.
You can try to perform the division of $1$ by $1+X$ to check that
$$(1+X)^{-1}=1-X+X^2-X^3+\dotsm$$
A: The Taylor series of $\frac{\sin z}{z}$ at the origin is straightforward to compute, the Taylor series of the reciprocal function $\frac{z}{\sin z}$ a bit less. However, we may notice that $\frac{z}{\sin z}$ is an even meromorphic function with simple poles at the elements of $\pi\mathbb{Z}\setminus\{0\}$. Additionally
$$ z\prod_{n\geq 1}\frac{1-\frac{z^2}{n^2 \pi^2}}{1-\frac{4z^2}{(2n-1)^2\pi^2}} =\tan(z)\tag{1}$$
leads to:
$$ \frac{1}{\sin z}=\frac{d}{dz}\log\tan\frac{z}{2}\\=\frac{1}{z}+\sum_{n\geq 1}\left(\frac{1}{\pi -2 n \pi -z}+\frac{1}{-2 n \pi +z}-\frac{1}{\pi -2 n \pi +z}+\frac{1}{2 n \pi +z}\right) \tag{2}$$
which simplifies into
$$ \frac{z}{\sin z}=1+2z^2\sum_{n\geq 1}\left(\frac{1}{\left[(2n-1)\pi\right]^2-z^2}-\frac{1}{\left[2n \pi\right]^2-z^2}\right)\tag{3}$$
then into:
$$\begin{eqnarray*} \frac{z}{\sin z}&=&1+2\sum_{n\geq 1}\sum_{h\geq 0}\frac{z^{2h+2}}{\pi^{2h+2}}\left(\frac{1}{(2n-1)^{2h+2}}-\frac{1}{(2n)^{2h+2}}\right)\\&=&1+\sum_{h\geq 0}\frac{z^{2h+2}}{\pi^{2h+2}}\zeta(2h+2)\left(2-\frac{1}{4^{h}}\right)\end{eqnarray*}\tag{4}$$
from which it is clear that the Taylor series of $\frac{z}{\sin z}$ at the origin only has non-negative coefficients.
That can be seen also as a consequence of $\frac{1}{\sin z}=\cot(z/2)-\cot(z)$ and the well-known identity
$$ \sum_{h\geq 1}\zeta(2h)z^{2h}=\frac{1-\pi z\cot(\pi z)}{2}.\tag{5} $$
