TAN.

Since $x_n \sim (n-\tfrac12)\pi$ as $n\to\infty$ , it is clear that $\sum_{n=1}^\infty x_n^{-2}$ converges. Now, since $f(z)=\frac{\sin z}{z}-\cos z$ is an even entire function of order $1$ with a double zero at $0$ , the Hadamard factorisation theorem tells us that $f(z)=Az^2\prod_{n=1}^\infty\left(1-\frac{z^2}{x_n^2}\right)$ But $f(z)=\tfrac13z^2-\tfrac{1}{30}z^4+\mathrm{O}(z^6)$ So it follows that. $A=\tfrac13$ and $S^{-1}=\boxed{10}$ .