<s>[INST] summation of a trigonometric series How to evaluate $\tan^2(1) + \tan^2(3) + \tan^2(5) + \tan^2(7) + \ldots + \tan^2(89)$? 
Angles are given in degrees.
I tried converting $\tan(89)$ as $\cot(1)$ and then tried combining $\tan(1)$ and $\cot(1)$, but later got stuck.
 [/INST] For integer $n,\tan(2n+1)90^\circ=\tan90^\circ=\infty$
If $\tan90x=\infty,90x=180^\circ m+90^\circ=90^\circ(2m+1)$ where $m$ is any integer
$\implies x=(2m+1)^\circ$ where $0\le m\le89$
Like Trig sum: $\tan ^21^\circ+\tan ^22^\circ+...+\tan^2 89^\circ = ?$, 
$$\tan90x=\frac{\binom{90}1t-\binom{90}3t^3+\cdots+\binom{90}{89}t^{89}}{\binom{90}0-\binom{90}2t^2+\cdots-\binom{90}{90}t^{90}}$$ where $t=\tan x$
For $\tan90x=\infty,$  $$\binom{90}0-\binom{90}2t^2+\cdots+\binom{90}{88}t^{88}-\binom{90}{90}t^{90}=0 \iff t^{90}-\binom{90}{88}t^{88}+\cdots-1=0$$
Now as $\tan(180^\circ-y)=-\tan y\implies\tan^2(180^\circ-y)=\tan^2y,$
So, the roots of $$u^{45}-\binom{90}{88}u^{44}+\cdots-1=0$$ are $\tan(2r+1)^\circ,0\le r\le 44$ or $45\le r\le89$
$$\implies\sum_{r=0}^{44}\tan^2(2r+1)^\circ=\frac{\binom{90}{88}}1=\binom{90}2=\cdots$$
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