Infinite sum of reciprocals of squares of lengths of tangents from origin to the curve $y=\sin x$ 
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.
I could not think of any method so any other method would be openly welcomed. 
Any help would be very beneficial to solve this problem.  
Thanks in advance. 

Edit:
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 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  Wolfy 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: Answer to original question
Simple bound $\pi k\leq a_k \leq \sqrt{\pi^2(k+\frac12)^2+1}$ shows that $\dfrac{a_k}{a_{k+1}}\to 1$.
So both sums diverge.
Answer to modified question
Again, (1) diverges.
(2) also diverges since squaring the ratio doesn't change $\to 1$.
(3) converges since you have $\pi k<\lambda_k=x_k<a_k$ giving $\dfrac{1}{a_k^2}\leq\dfrac{1}{\lambda_k^2}\leq\dfrac{1}{k^2}$.  This is of course a very loose bound.
Finding $\displaystyle\sum_{k=1}^\infty\frac{1}{\lambda_k^2+2}$ for use in $\sum a_k^{-2}$
Recall one way of finding $\displaystyle\sum_{k=1}^\infty\lambda_k^{-2}=\frac{1}{10}$ is write down the series expansion of
$$
\sin x-x\cos x
$$
and set that to zero, reading off the lowest terms
$$
x^3\left(\frac{1}{3}-\frac{x^2}{30}+\frac{x^4}{840}+\dots\right)=0
$$
and after cancelling $x^3$ factor in front, you read off $\dfrac{1/30}{1/3}$ in a reciprocal Viete's formula way (except to make it rigourous you need to do it properly with infinite products, but that's another story).
So now we want to do this with $\lambda_k^2+2$.  You want to construct a series whose roots are $\lambda_k^2+2$  The simplest heuristic way is to use the full series expansion above (ignoring the $x^3$) and try to express it as a power series in $x^2+2$, and read off the sum of reciprocals of roots.
