<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] My answer is based on a rather beautiful answer given to a related question, that can be found here.
Expressing the summation in radians, we wish to evaluate
$$S=\sum_{m=0}^{45-1}\tan^2\left\{\frac{(2m+1)\pi}{2\cdot90}\right\}=\sum_{m=0}^{45-1}\tan^2\left\{\frac{(m+\frac{1}{2})\pi}{2\cdot45}\right\}$$
Using the well-known identity that
$$\left(\cos{\frac{k\pi}{2n}} + i\sin{\frac{k\pi}{2n}}\right)^{2n}=(-1)^{k}$$ and by setting $k=(m+\frac{1}{2})$ we have a purely imaginary expression
$$\left(\cos{\frac{(m+\frac{1}{2})\pi}{2n}} + i\sin{\frac{(m+\frac{1}{2})\pi}{2n}}\right)^{2n}=(-1)^{m}i$$ 
Calculating the binomial expansion, where $n=45$, we have
$$\sum_{t=0}^{2n}{2n \choose t}\left[\cos\frac{(m+\frac{1}{2})\pi}{2n}\right]^t\left[i\sin\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2n-t}$$
Considering only the real terms, which should be $0$, we have
$$\sum_{r=0}^{n}{2n \choose 2r}\left[\cos\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2r}(-1)^{n-r}\left[\sin\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2n-2r}=0$$
dividing by $\left[\cos\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2n}$ results in
$$\sum_{r=0}^{n}{2n \choose 2r}(-1)^{n-r}\left[\tan\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2n-2r}=0$$
In a manner similar to the linked answer, we use Vieta's formula, as the summation we wish to evaluate is simply the sum of the roots of the polynomial. Thus, the sum
is ($n=45$)
$$S=\frac{{2n \choose 2}}{{2n \choose 0}}=\frac{(2n)(2n-1)}{2}=\frac{90\cdot89}{2}=4005$$
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