Solution
= Solution
The <Taylor series> $\sin x=x-x^3/6+O(x^5)$ gives $\sin^2x=x^2(1-x^2/3+O(x^4))$. Taking its reciprocal and cancelling the singular term gives
$$
\frac1{\sin^2x}-\frac1{x^2}=\frac1{x^2}\left[1+\frac{x^2}{3}+O(x^4)-1\right]=\boxed{\frac13+O(x^2)}.
$$
Thus the apparent singularity at zero is removable, and the first nonzero term is the constant $1/3$.