Bromance <3

Notice that $\cot\left(\frac\pi2-x\right)=\tan x$ . Hence,

$\tan^8\frac{\pi}{24}+\tan^8\frac{5\pi}{24}+\tan^8\frac{7\pi}{24}+\tan^8\frac{11\pi}{24}=\tan^8\frac{\pi}{24}+\tan^8\frac{5\pi}{24}+\cot^8\frac{7\pi}{24}+\cot^8\frac{11\pi}{24}$

Letting $x=\tan\frac\pi{24}$ and $y=\tan\frac{5\pi}{24}$ , we get $x^8+\frac1{x^8}+y^8+\frac1{y^8}$ . But

$x^8+\frac1{x^8}=\left(x^4+\frac1{x^4}\right)^2-2=\left(\left(x^2+\frac1{x^2}\right)^2-2\right)^2-2=\left(\left(\left(x+\frac1x\right)^2-2\right)^2-2\right)^2-2$

Similarly,

$y^8+\frac1{y^8}=\left(\left(\left(y+\frac1y\right)^2-2\right)^2-2\right)^2-2$

Hence, it suffices to find $x+\frac1x$ and $y+\frac1y$ . So $\tan\frac\pi{24}+\frac1{\tan\frac\pi{24}}=\frac1{\sin\frac{\pi}{24}\cos\frac\pi{24}}=\frac{2}{\sin\frac{\pi}{12}}$ . Similarly, $\tan\frac{5\pi}{24}+\frac1{\tan\frac{5\pi}{24}}=\frac{2}{\sin\frac{5\pi}{12}}$ . But $\sin\frac{\pi}{12}=\frac{\sqrt6-\sqrt2}4$ and $\sin\frac{5\pi}{12}=\frac{\sqrt6+\sqrt2}4$ so $x+\frac1x=2\left(\sqrt6+\sqrt2\right)$ and $y+\frac1y=2\left(\sqrt6-\sqrt2\right)$ . Using the formulas above, we can easily compute that $x^8+\frac1x^8+y^8+\frac1y^8=11080708$ .

If he would have known this method, then definitely he would have posted a better solution than mine. But thanks! Cody Johnson. :D

yes, u r right....

Why are you posting this everywhere?

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