Find $\quad\cot^2(2^{\circ})+\cot^2(6^{\circ})+\cot^2({10}^{\circ})++\cdots +\cot^2({86}^{\circ})$

Find $$\cot^2(2^{\circ})+\cot^2(6^{\circ})+\cot^2({10}^{\circ})++\cdots +\cot^2({86}^{\circ})$$

$\mathbf {My Attempt}$
I tried to write the sum backward like this
$$S=\sum_{n=1}^{22} \cot^2(4n-2) = \sum_{n=1}^{22} \cot^2({90}^{\circ}-4n) = \sum_{n=1}^{22} \tan^2(4n)$$ $S=990$ But still can't find any good trigonometry identity to form telescopic series or something like that.
Any hint?

• – lab bhattacharjee Jul 18 '18 at 1:35
• @lab many thx. I seem I've overlooked this post. I can adapt it now. – Wolfdale Jul 18 '18 at 1:59