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Question Number 102910 by bramlex last updated on 11/Jul/20

$$\left(1\right)\underset{\mathrm{m}=1}{\sum ^{\mathrm{n}}}\tan {}^2\left(\frac{\mathrm{m}\pi }{2\mathrm{n}+1}\right)$$$$\left(2\right)\underset{\mathrm{m}=1}{\prod ^{\mathrm{n}}}\tan {}^2\left(\frac{\mathrm{m}\pi }{2\mathrm{n}+1}\right)$$

Answered by floor(10²Eta[1]) last updated on 11/Jul/20

$$\mathrm{go}\mathrm{see}\mathrm{michal}\mathrm{penn}\mathrm{channel}\mathrm{on}\mathrm{yt}$$$$$$

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