Expand in Taylor series $\frac{1}{1-\sin{x}}$ Expand in Taylor series $\frac{1}{1-\sin{x}}$
I have an idea that $\frac{1}{1-\sin{x}} = 1 + \sin {x} + \sin^2 {x} + \sin^3 {x} + \dots$
But I don't know what to do next. Every sine expands in infinity series...
Can anybody help?

$$\sin^n(x)=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^n=\frac1{(2i)^n}\sum_{k=0}^n\binom nke^{i(2k-n)x}=\\
\frac1{(2i)^n}\sum_{k=0}^n\binom nk\sum_{j=0}^\infty\frac{i^j(2k-n)^j}{j!}x^j,$$
and
$$\frac1{1-\sin(x)}=\sum_{n=0}^\infty\frac1{(2i)^n}\sum_{k=0}^n\binom nk\sum_{j=0}^\infty\frac{i^j(2k-n)^j}{j!}x^j.$$
Not the simplest on Earth, but usable for the first few terms.
