Question Number 102910 by bramlex last updated on 11/Jul/20
$$\left(\mathrm{1}\right)\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{m}\pi}{\mathrm{2n}+\mathrm{1}}\right) \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{m}\pi}{\mathrm{2n}+\mathrm{1}}\right)\: \\ $$
Answered by floor(10²Eta[1]) last updated on 11/Jul/20
$$\mathrm{go}\:\mathrm{see}\:\mathrm{michal}\:\mathrm{penn}\:\mathrm{channel}\:\mathrm{on}\:\mathrm{yt} \\ $$$$ \\ $$