Answer
$$1-x+x^2-\dfrac{5x^3}{6}+...$$
Work Step by Step
We have the Maclaurin series for $\dfrac{1}{ (1-x)}$ as follows
$$\dfrac{1}{ (1-x)}=1+x+x^2+x^3+...$$
Replace $\sin x$ for $x$ in the above form.
Therefore, we have:
$$\dfrac{1}{1+\sin x}=1-\sin x+\sin^2 x +...
\\=1-(x-\dfrac{x^3}{6}) +(x-\dfrac{x^3}{6})^2 +...\\
=1-x+x^2-\dfrac{5x^3}{6}+...$$
So, the four terms of the Maclaurin series for $f(x)=\dfrac{1}{1+\sin x}$ are: $$1-x+x^2-\dfrac{5x^3}{6}+...$$
