Summing tangent slopes for trig function fixed points Let's consider the values of $x$ for which the tangent line to the graphs of $\sin x$, $\cos x$, $\csc x$, and $\sec x$ passes through the origin. These values of $x$ satisfy $x=\tan x$, $-x=\cot x$, $-x=\tan x$, and $x=\cot x$, respectively. 
The values of $m$ for which $y=mx$ is tangent to $y=\sin x$ are $1$, $-0.217234$, $0.128375$, $-0.0913252$, $0.0709135$, $-0.0579718$, $0.0490296$, $-0.0424796$, et cetera. 
Let $a$ denote the sum of all possible values of $m$ for positive $x$ values giving tangency. Thus, $a=1+\displaystyle\sum_{x=\tan x;\,x>0}\frac{\sin x}{x}$.
Similarly, the values of $m$ for which $y=mx$ is tangent to $y=\cos x$ are $-0.336508$, $0.161228$, $-0.106708$, $0.0798312$, $-0.0637916$, $0.0531265$, et cetera.
Similarly, let $b=\displaystyle\sum_{-x=\cot x;\,x>0}\frac{\cos x}{x}$,
$c=\displaystyle\sum_{-x=\tan x;\,x>0}\frac{\csc x}{x}$,
And $d=\displaystyle\sum_{x=\cot x;\,x>0}\frac{\sec x}{x}$.
Where $x$ in all of these sums is in radians.
My question is, what values do each of these sums ($a$, $b$, $c$, and $d$) converge to? Are they "nice" values in that they can be easily expressed? Or do they at least have closed-form expressions?
 A: I don't know about the values but it would be interesting to see that these sums converge:
We know tangent is periodic: $\tan|_{]-\frac{\pi}{2}, \frac{\pi}{2}[} = \tan|_{]k\pi-\frac{\pi}{2}, k\pi+\frac{\pi}{2}[}$.
Let's show that tan has exactly one fixed point on each $I_k := ]k\pi-\frac{\pi}{2}, k\pi+\frac{\pi}{2}[$. 
At least one: Define the coninous function $f(x) := \tan(x)-x$. 
Since $\lim\limits_{x \to k\pi-\frac{\pi}{2}} f(x) = -\infty$ and $\lim\limits_{x \to k\pi-\frac{\pi}{2}} f(x) = \infty$ and tan is continous, we know that there exists a $p$ such that $f(p)=0 \Rightarrow tan(p)=p$.
At most one: We know $\frac{d}{dx}\tan(x) = \sec^2(x)$ and $\sec^2(x) = 1 \Leftrightarrow x \in \pi\mathbb{Z}$. If $\tan$ had two fixed-points a,b with $a \neq b$ in $I_k$, then, by meanvalue theorem, $\frac{\tan(b)-\tan(a)}{b-a} = \frac{b-a}{b-a} = sec^2(x)$ for some $x \in I_k$. This contradicts $sec^2(x) = 1 \Leftrightarrow x \in \pi\mathbb{Z}$. Therefore the fixed-point is unique in $I_k$. Let's call it $x_k$.
Since $\tan(x)$ is strictly increasing on $I_k$, we know that for bigger x our fixed point $x_k \in I_k$ moves right in $I_k$.
Since $x_k$ tends towards a multiple of $\pi$ plus $\frac{\pi}{2}$, $\cos(x_k) \to 0$ for $k \to \infty$.
Furthermore $\cos(x) > 0\ \forall x \in I_k$ if k is even and $\cos(x) < 0\ \forall x \in I_k$ if k is odd, and therefore we know that
$$\sum\limits_{x=tan(x), x > 0} \frac{\sin(x)}{x} = \sum\limits_{k = 1}^\infty \frac{\sin(x_k)}{x_k} = \sum\limits_{k = 1}^\infty \frac{\sin(x_k)}{\tan(x_k)} = \sum\limits_{k = 1}^\infty \cos(x_k) = \sum\limits_{k = 1}^\infty (-1)^k |\cos(x_k)| < \infty$$
by the alternating series test.
Please correct me if something is wrong. These were my first thoughts on the problem and I thought I'd share them.
