How do you find the value of $\sum_{r=0}^{44} \tan^2(2r+1)$?

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?

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 $$ \begin{align} \sum_{k=0}^{44}\tan^2\left(\frac{(2k+1)\pi}{180}\right) &=\frac{15931}{3}-\frac{3916}{3}\\ &=4005 \end{align} $$