For the sake of at least having an answer, here's what's in J.M.'s link, in a nutshell anyway.
Note that $\tan(x)-x$ and $\sin(x)-x\cos(x)$ have the same zeros but the latter is holomorphic on the complex plane - the difference is that the latter has a triple root at $0$ whereas the former only has a double root, but our sum doesn't involve $x=0$ anyway. This means it affords a Weierstrass factorization. Some work can show that the coefficients of the expanded-out polynomials of the partial products in such a factorization will, in the limit, converge to those of the Taylor expansion of the function so long as we ignore the $e^{g(z)}$ and $E_n(z)$ factors. We can also show that the zeros are approximately $x_n\approx(n+\frac{1}{2})\pi$ asymptotically, and comparing this with the factorization for, say, $\sin$ tells us we don't need any $E_n$ factors. Now putting together the series expansions for sine and cosine gives
$\left(x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\cdots\right)-x\left(1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\cdots\right)$
$=\frac{1}{3}x^3\left(1-\frac{1}{10}x^2+\cdots\right).$
We ignore the $x^3$ and normalize so that $a_0=1,a_1=0,a_2=-1/10$. Now compute
$\sum_{\tan(u)=u\ne0}u^{-2}=\left(\sum_{i}\frac{1}{\lambda_i}\right)^2-2\left(\sum_{i
$=a_1^2-2a_2=\frac{1}{5}.$
Note that $\tan(x)-x$ is an odd function and squaring takes out signs so by symmetry the above sum essentially double-counts every positive root. Divide by $2$ and our final answer is $1/10$.
Note that all sums-of-reciprocals of even powers of solutions to $\tan(x)=x$ can be evaluated in this way by using the Newton-Girard formulas.