Infinite sum of reciprocals of squares of lengths of tangents from origin to the curve $y=\sin x$ This question is actually from MSE. I had to post it here due to the lack of response there even after placing a bounty. Here goes the question 

Let tangents be drawn to the curve $y=\sin x$ from the origin. Let the points of contact of these tangents with the curve be $(x_k,y_k)$ where $x_k\gt 0 ;(k\ge 1)$ such that $x_k\in (\pi k,  (k+1)\pi)$ and $$a_k=\sqrt {x_k^2+y_k^2}$$ (Which is basically the distance between the corresponding point of contact and the origin i.e. the length of tangent from origin)  . 


I wanted to know the value of  

$$\sum_{k=1}^{\infty} \frac {1}{a_k ^2}$$

Now this question has just popped out in my brain and is not copied from any assignment or any book so I don't know whether it will finally reach a conclusion or not. 

I tried writing the equation of tangent to this curve from origin and then finding the points of contact but did not get a proper result which just that the $x$ coordinates of the points of contact will be the positive solutions of the equation $\tan x=x$ 
On searching internet for sometime about the solutions of $\tan x=x$ I got two important properties of this equation.  If $(\lambda _n)_{n\in N}$ denote the roots of this equation then 

$$1)\sum_n^{\infty} \lambda _n \to \infty$$ $$2)\sum_n^{\infty} \frac {1}{\lambda _n^2} =\frac {1}{10}$$

But were not of much help. 
I also tried writing the points in polar coordinates to see if that could be of some help but I still failed miserably.

On trying a bit more using some coordinate geometry I found that the locus of the points of contact is $$x^2-y^2=x^2y^2$$
Hence for third sum we just need to find $$\sum_{k=1}^{\infty} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{2\lambda _k ^2} +\sum_{k=1}^{\infty} \frac {1}{2(\lambda _k ^2 +2)} =\frac {1}{20}+\frac {1}{2}\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 +2} 
 $$
Now for the second summation I did think about it to form a series but for the roots to be $\lambda _k^2 +2$ we just need to substitute $x\to \sqrt {x−2}$ in power series of $\frac {\sin x-x\cos x}{x^3}$ and then get the result but it was still a lot confusing for me. 
Using $x\to\sqrt {x-2}$ in the above power series and using  Wolfram Alpha I have got a series. So we need ratio of coefficient of $x$ to the constant term so is the value of second summation equal to
$$\frac {5\sqrt 2\sinh(\sqrt 2)−6\cosh(\sqrt 2)}{4(2\cosh(\sqrt 2)−\sqrt 2\sinh(\sqrt 2))}?$$
Is this value correct or did I do it wrong? 
I would also like to know if there is some other method to solve this problem
 A: It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer, and as I understand the claim there this series equals $\frac{1}{3}\prod_{k\geq 1} (1-\frac{x^2}{\lambda_k^2})$. I have not verified this claim, but if it's true, then your approach does make sense. UPD. Precise argument can be seen in this answer.
I've also verified your analytical expression numerically, and it looks fine. Namely, the analytical expression gives $\sum_k a_k^{-2} = 0.0973745978\ldots$, while numerically we have $\sum_{k=1}^{10^5} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)} = 0.09737358469\ldots$ with the difference $\approx 10^{-6}$.
A: One perhaps more direct way could be to use your expression $\frac {1}{20}+\frac {1}{2}\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 +2}$, expand it ($\frac {1}{20}+\frac {1}{2}\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2} (1-\frac{2}{\lambda_k^2} + \cdots)$) and calculate this sum by using the sums-of-reciprocals of even powers of $\lambda_k$ which can be calculated in a simliar way as the sum of reciprocal squares of $\lambda_k$ (see the last sentence of the accepted answer of 
https://math.stackexchange.com/questions/75206/sum-of-the-squares-of-the-reciprocals-of-the-fixed-points-of-the-tangent-functio). 
