I have to find the first 4 terms of $\frac{x}{\sin x}$. My first step was to calculate the first 4 terms of the Maclaurin series for $\sin x$ which are $$\frac{x}{1!} – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!}$$
Then, we will have that the first 4 terms of $\frac{x}{\sin x}$ are: $$ \frac{x}{\frac{x}{1!} – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!}}$$
which would give me $$1 – \frac{3!}{x^2} + \frac{5!}{x^4} – \frac{7!}{x^6}$$ however that answer is incorrect. My question is: why can't we just divide by $x$ to get the answer? When we do the Maclaurin series for $\frac{\sin x}{x}$ we just divide by $x$ (see attached picture below) 
Why does this same concept not work when getting the Maclaurin series for $\frac{x}{\sin x}$?
Best Answer
Notice that when you divide by more one term you can not divide term by term.
For example $$ \frac {1}{2+3} \ne \frac {1}{2}+\frac {1}{3}$$
Thus dividing $x$ by $\sin x$ requires more than just dividing term by term.
You may assume that $$\frac {x}{\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+...} =a_0+a_1x +a_2x^2 +...$$ and try to find the coefficients by cross multiplication.