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?