# Value of tan 2 </msup> &#x2061;<!-- ⁡ --> ( 5 ) + tan

Value of ${\mathrm{tan}}^{2}\left(5\right)+{\mathrm{tan}}^{2}\left(10\right)+{\mathrm{tan}}^{2}\left(15\right)+...+{\mathrm{tan}}^{2}\left(85\right)$
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behk0
${\mathrm{tan}}^{2}\left({5}^{\circ }\right)+{\mathrm{tan}}^{2}\left({10}^{\circ }\right)+...+{\mathrm{tan}}^{2}\left({85}^{\circ }\right)={\mathrm{tan}}^{2}\left(\frac{5\pi }{180}\right)+{\mathrm{tan}}^{2}\left(\frac{10\pi }{180}\right)+...+{\mathrm{tan}}^{2}\left(\frac{85\pi }{180}\right)$
$=\sum _{r=1}^{17}{\mathrm{tan}}^{2}\left(\frac{r\pi }{2\cdot 18}\right)$
Now note this result: Prove that $\sum _{k=1}^{n-1}{\mathrm{tan}}^{2}\frac{k\pi }{2n}=\frac{\left(n-1\right)\left(2n-1\right)}{3}$
We have n=18 so then we get:
$S=\frac{\left(18-1\right)\left(2\cdot 18-1\right)}{3}=\frac{595}{3}$