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.
You can use Maclaurin's expansion to expand in powers of x as follows:


$f(x)$=$f(0)$+$xf'(0)$+$(x^2$/$2!$)$f''(0)$+...


where $f(x)=1/(1-sinx )$