Good Bye 2015 - 1

$\begin{aligned}\begin{aligned}\dfrac{3-\tan^2{\left(\frac{\pi}7\right)}}{1-\tan^2{\left(\frac{\pi}7\right)}}\div\cos\left(\frac{\pi}7\right)=&\ \dfrac1{\cos\left(\frac{\pi}7\right)}\left(\dfrac{3-\tan^2{\left(\frac{\pi}7\right)}}{1-\tan^2{\left(\frac{\pi}7\right)}}\right)\\=&\ \sec\left(\frac{\pi}7\right)\left(\dfrac{\left(1-\tan^2{\left(\frac{\pi}7\right)}\right)+2}{1-\tan^2{\left(\frac{\pi}7\right)}}\right)\\=&\ \sec\left(\frac{\pi}7\right)\left(1+\dfrac2{1-\dfrac{\sin^2{\left(\frac{\pi}7\right)}}{\cos^2{\left(\frac{\pi}7\right)}}}\right)\left(\dfrac{\cos^2\left(\frac{\pi}7\right)}{\cos^2\left(\frac{\pi}7\right)}\right)\\=&\ \sec\left(\frac{\pi}7\right)\left(1+\dfrac{2\cos^2{\left(\frac{\pi}7\right)}}{\cos^2{\left(\frac{\pi}7\right)}-\sin^2{\left(\frac{\pi}7\right)}}\right)\\=&\ \sec\left(\frac{\pi}7\right)\left(1+\dfrac{2\cos^2{\left(\frac{\pi}7\right)}}{2\cos^2{\left(\frac{\pi}7\right)}-1}\right)\\=&\ \sec\left(\frac{\pi}7\right)\left(1+\dfrac{\left(2\cos^2{\left(\frac{\pi}7\right)}-1\right)+1}{2\cos^2{\left(\frac{\pi}7\right)}-1}\right)\\=&\ \sec\left(\frac{\pi}7\right)\left(2+\dfrac1{\cos{\left(\frac{2\pi}7\right)}}\right)\\=&\ 2\sec{\left(\frac{\pi}7\right)}+\sec{\left(\frac{\pi}7\right)}\sec{\left(\frac{2\pi}7\right)}\end{aligned}\end{aligned}$

Let $A=2\sec{\left(\frac{\pi}7\right)}+\sec{\left(\frac{\pi}7\right)}\sec{\left(\frac{2\pi}7\right)}$

Multiplying both sides with $\cos{\left(\frac{\pi}7\right)}\cos{\left(\frac{2\pi}7\right)}$

$\begin{aligned}\begin{aligned}A\cos{\left(\frac{\pi}7\right)}\cos{\left(\frac{2\pi}7\right)}=&\ 2\sec{\left(\frac{\pi}7\right)}\cos{\left(\frac{\pi}7\right)}\cos{\left(\frac{2\pi}7\right)}+\sec{\left(\frac{\pi}7\right)}\sec{\left(\frac{2\pi}7\right)}\cos{\left(\frac{\pi}7\right)}\cos{\left(\frac{2\pi}7\right)}\\A\cos{\left(\frac{\pi}7\right)}\cos{\left(\frac{2\pi}7\right)}=&\ 2\cos{\left(\frac{2\pi}7\right)}+1\\A\cos{\left(\frac{\pi}7\right)}\cos{\left(\frac{2\pi}7\right)}=&\ 2\left(2\cos^2{\left(\frac{\pi}7\right)}-1\right)+1\\A\cos{\left(\frac{\pi}7\right)}\cos{\left(\frac{2\pi}7\right)}=&\ 4\cos^2{\left(\frac{\pi}7\right)}-1\end{aligned}\end{aligned}$

Multiplying both sides with $4\sin{\left(\frac{\pi}7\right)}$

$\begin{aligned}\begin{aligned}4A\cos{\left(\frac{\pi}7\right)}\cos{\left(\frac{2\pi}7\right)}\sin{\left(\frac{\pi}7\right)}=&\ 4\left(4\cos^2{\left(\frac{\pi}7\right)}-1\right)\left(\sin{\left(\frac{\pi}7\right)}\right)\\2A\left(2\sin{\left(\frac{\pi}7\right)}\cos{\left(\frac{\pi}7\right)}\right)\cos{\left(\frac{2\pi}7\right)}=&\ 4\sin{\left(\frac{3\pi}7\right)}\\A\left(2\sin{\left(\frac{2\pi}7\right)}\cos{\left(\frac{2\pi}7\right)}\right)=&\ 4\sin{\left(\pi-\frac{3\pi}7\right)}\\A\sin{\left(\frac{4\pi}7\right)}=&\ 4\sin{\left(\frac{4\pi}7\right)}\\A=&\ 4\end{aligned}\end{aligned}$

Thus, $\dfrac{3-\tan^2{\left(\frac{\pi}7\right)}}{1-\tan^2{\left(\frac{\pi}7\right)}}\div\cos\left(\dfrac{\pi}7\right)=\boxed4$

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Notes:

$\dfrac1{\cos{\theta}}=\sec{\theta}$

$\tan^2{\theta}=\dfrac{\sin^2{\theta}}{\cos^2{\theta}}$

$\sin{2\theta}=2\sin{\theta}\cos{\theta}$

$\sin{3\theta}=\sin{\theta}(4\cos^2{\theta}-1)$

$\cos{2\theta}=\cos^2{\theta}-\sin^2{\theta}=2\cos^2{\theta}-1$

Let $w=e^{2 \pi i/7}$ . We have the following equations: $\begin{aligned} w^6+w^5+\cdots+w+1&=0\\ \tan\left(\dfrac{\pi}{7}\right)&=\dfrac{w-1}{i(w+1)}\\ \cos\left(\dfrac{\pi}{7}\right)&=\dfrac{1}{2}\lvert 1+w \rvert \end{aligned}$

Let the expression be $S$ . Note that $S>0$ , then we have: $\begin{aligned} S&=\dfrac{3-\left(\frac{w-1}{i(w+1)}\right)^2}{1-\left(\frac{w-1}{i(w+1)}\right)^2} \cdot \dfrac{1}{\frac{1}{2}\lvert 1+w \rvert}\\ &=\dfrac{3(w+1)^2+(w-1)^2}{(w+1)^2+(w-1)^2} \cdot \dfrac{2}{\lvert 1+w \rvert}\\ &=\dfrac{4w^2+4w+4}{2w^2+2} \cdot \dfrac{2}{\lvert 1+w \rvert}\\ &=4\left(\dfrac{w^2+w+1}{(w^2+1) \lvert 1+w \rvert}\right)\\ &=4\left\lvert \dfrac{w^2+w+1}{(w^2+1)(1+w)} \right\rvert\\ &=4\left\lvert \dfrac{w^2+w+1}{w^3+w^2+w+1} \right\rvert\\ &=4\left\lvert \dfrac{w^2+w+1}{-w^4-w^5-w^6} \right\rvert\\ &=4\left\lvert \dfrac{w^2+w+1}{-w^4(w^2+w+1)} \right\rvert\\ &=4\left\lvert \dfrac{w^2+w+1}{w^2+w+1} \right\rvert\\ &=\boxed{4} \end{aligned}$