Prove that $\sum_{k=1}^{89} \frac{1}{\tan^3(k^\circ)+1}=44.5$ So I've been stuck on this problem for a while now, and to be honest I have no idea how to start. I stumbled upon it while practicing for an Olympiad, and it is a past Olympiad problem (for teens) itself.
The problem is as follows:
Given is the function $f(x) = \frac{1}{x^3+1}$
Prove that:
$$\sum_{k=1}^{89} f(\tan(k^\circ)) = 44.5$$
In other words, prove that:
$$\frac{1}{\tan^3(1^\circ)+1} + \frac{1}{\tan^3(2^\circ)+1} + \cdots + \frac{1}{\tan^3(89^\circ)+1} = 44.5 $$
The question was multiple choice at first, but I managed to figure out 44.5 with my calculator. I still wondered how one would come up with a proof.
Any guidelines or straight up solutions are welcome!

We can write $\tan(90^\circ - k^\circ) = \frac{1}{\tan(k^\circ)}$.
\begin{align*}
\sum_{k=1}^{89} \frac{1}{\tan^3(k^\circ) + 1} &= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=46}^{89} \frac{1}{\tan^3(k^\circ) + 1} + \frac{1}{\tan^3(45^\circ) + 1} \\
&= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=1}^{44} \frac{1}{\tan^3(90 - k^\circ) + 1} + \frac{1}{2} \\
&= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=1}^{44} \frac{1}{\frac{1}{\tan^3(k^\circ)} + 1} + \frac{1}{2} \\
&= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=1}^{44} \frac{\tan^3(k^\circ)}{\tan^3(k^\circ) + 1} + \frac{1}{2} \\
&= \sum_{k=1}^{44} \frac{\tan^3(k^\circ) + 1}{\tan^3(k^\circ) + 1} + \frac{1}{2} \\
&= 44.5
\end{align*}
