How do you find the value of $\sum_{r=0}^{44} \tan^2(2r+1)$? Problem: 
Find the value of $$\sum_{r=0}^{44} \tan^2(2r+1)$$
Note: The angles here are in degrees.

I don't know how to solve this question because trigonometric simplifications didn't get me anywhere. I think there was a method to solve this question using complex numbers which I no longer remember. Any hint/help will be appreciated.
 A: Using Sum of tangent functions where arguments are in specific arithmetic series,
$$\tan90x=\dfrac{\binom{90}1t-\binom{90}3t^3+\cdots+\binom{90}{89}t^{89}}{\binom{90}0-\binom{90}2t^2+\cdots+\binom{90}{90}t^{90}}$$ where $t=\tan x$ 
If $\tan 90x=\tan90^\circ=\infty$
$90x=180^\circ n+90^\circ=90^\circ(2n+1)$ where $n$ is any integer
$\implies x=(2n+1)^\circ$ where $n\equiv0,1,2,\cdots,88,89\pmod{90}$
So, the roots of $$t^{90}-\binom{90}{88}t^{88}+\cdots=0$$ 
are  $\tan(2n+1)^\circ$ where $n\equiv0,1,2,\cdots,88,89\pmod{90}$
Now as $-\tan(2n+1)^\circ=\tan\{180^\circ-(2n+1)^\circ\}=\tan\{2(89-n)+1)^\circ\}$
$$\implies\sum_{r=0}^{89}\tan^2(2n+1)^\circ=2\sum_{r=0}^{44}\tan^2(2n+1)^\circ$$
and the roots of $$s^{45}-\binom{90}{88}s^{44}+\cdots=0$$ are $\tan^2(2n+1)^\circ$ where $n\equiv0,1,2,\cdots,88,89\pmod{90}$
Using Vieta's formula,
$$\sum_{r=0}^{89}\tan^2(2n+1)^\circ=\binom{90}{88}=\binom{90}{90-88}=?$$
Can you take it from here?
A: Using an idea similar to this answer, note that the function
$$
\frac{90/z}{z^{90}-1}
$$
has residue $1$ for $z=e^{k\pi i/45}$ and residue $-90$ at $z=0$.
On $|z|=1$,
$$
\tan(\theta/2)=-i\frac{z-1}{z+1}
$$
Integrating
$$
f(z)=-\left(\frac{z-1}{z+1}\right)^2\frac{90/z}{z^{90}-1}
$$
around a large circle is $0$ since the integrand is approximately $|z|^{-91}$. Thus, the sum of residues is
$$
2\sum_{k=0}^{44}\tan^2\left(\frac{2k\pi}{180}\right)+\operatorname*{Res}_{z=0}f(z)+\operatorname*{Res}_{z=-1}f(z)=0
$$
Since $\operatorname*{Res}\limits_{z=0}f(z)=90$ and $\operatorname*{Res}
\limits_{z=-1}f(z)=-\frac{8102}{3}$, we get
$$
\begin{align}
\sum_{k=0}^{44}\tan^2\left(\frac{2k\pi}{180}\right)
&=\frac12\left(\frac{8102}{3}-90\right)\\
&=\frac{3916}{3}
\end{align}
$$
The sum in the question is the difference between the value in answer cited above and this value, that is
$$
\begin{align}
\sum_{k=0}^{44}\tan^2\left(\frac{(2k+1)\pi}{180}\right)
&=\frac{15931}{3}-\frac{3916}{3}\\
&=4005
\end{align}
$$
