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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.