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Question Number 142393 by qaz last updated on 31/May/21

$$\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}-1}}{\sec }^2\left(\frac{\mathrm{k}\pi }{\mathrm{n}}\right)={\mathrm{n}}^2......???$$

Answered by mindispower last updated on 31/May/21

$$=\underset{k=0}{\sum ^{n-1}}(1+{tg}^2\left(\frac{k\pi }{n}\right))=n+S$$$$S=\underset{k=0}{\sum ^{n-1}}{tg}^2\left(\frac{k\pi }{n}\right)=\underset{k=1}{\sum ^{n-1}}\frac{1}{{cot}^2\left(\frac{k\pi }{n}\right)}withek\ne \frac{n}{2},n\notin 2\mathbb{N}$$$$n=2m+1$$$$tg\left(x\right)=\frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}$$$$let{(Z-1)}^n-{(Z+1)}^n=P\left(Z\right)=0$$$$\Rightarrow \frac{Z-1}{Z+1}=e^{2ik\frac{\pi }{n}}$$$$Z=\frac{e^{\frac{2ik\pi }{n}}+1}{1-e^{\frac{2ik\pi }{n}}}=icot\left(\frac{k\pi }{n}\right),k\in [1,n-1]$$$$Wehave\frac{P^{\prime }}{P}=\underset{k=1}{\sum ^{n-1}}\frac{1}{X-icot\left(\frac{k\pi }{n}\right)}$$$$\Rightarrow \frac{{PP}^{″}-p{\left(x\right)}^{\prime 2}}{p^2\left(x\right)}=\underset{k=1}{\sum ^{n-1}}-\frac{1}{{(x-icot\left(\frac{k\pi }{n}\right))}^2}$$$$p\left(0\right)=-2$$$$p^{\prime }\left(0\right)=0,n=2m+1$$$$p^{″}\left(z\right)=n(n-1){(z-1)}^{n-2}-n(n-1){(z+1)}^{n-2}$$$$=-2n(n-1)$$$$\frac{4n(n-1)}{4}=\underset{k=1}{\sum ^{n-1}}\frac{1}{{cot}^2\left(\frac{k\pi }{n}\right)}=n^2-n$$$$\underset{k=0}{\sum ^{n-1}}{sec}^2\left(\frac{k\pi }{n}\right)=n+S=n+n^2-n=n^2$$$$\iff \underset{k=0}{\sum ^{n-1}}{sec}^2\left(\frac{k\pi }{n}\right)=n^2$$$$$$$$$$$$$$

Commented by qaz last updated on 31/May/21

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}\mathrm{power}$$

Commented by mindispower last updated on 01/Jun/21

$$pleasur$$

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