14 and 7 won't let you make it easy #RMM(0)

My approach A complete trigonometrical approach

We have $\zeta = \csc \dfrac{\pi}{14}-4\cos \dfrac{2\pi}{7}$ Note that $\csc \dfrac{\pi}{14}= \sin \left( \dfrac{\pi}{14}\right)^{-1}$ also notice that $\sin \dfrac{\pi}{14} = \cos \dfrac{3\pi}{7}$ Thus, above expression can be expressed as $\begin{aligned} \zeta & = \dfrac{1}{\sin \dfrac{\pi}{14}} -4\cos \dfrac{2\pi}{7}\\& =\dfrac{1}{\cos \dfrac{2\pi}{7}}\left(1 -4\cos \dfrac{2\pi}{7}\cdot \cos\dfrac{3\pi}{7}\right)\end{aligned}$ Recall that $\cos\dfrac{\pi}{7}\cdot \cos\dfrac{2\pi}{7}\cdot \cos\dfrac{3\pi}{7}=\dfrac{1}{8}$ Plugging back the above expression we have $\begin{aligned} \zeta & = 4\cos\dfrac{2\pi}{7}\left(2\cos\dfrac{\pi}{7} -1\right)=\dfrac{4}{2} =\boxed{2}\end{aligned}$

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Proof of the identity: Set $\begin{aligned} \zeta_0& = \cos\dfrac{2\pi}{7}\left(2\cos\dfrac{\pi}{2}-1\right) \\& =\dfrac{\sin\dfrac{4\pi}{7}}{2\sin\dfrac{2\pi}{7}}\left(2\cos\dfrac{\pi}{2}-1\right)\\& = \dfrac{1}{2\sin\dfrac{2\pi}{7}}\left(\sin\dfrac{5\pi}{7}+\sin\dfrac{3\pi}{7}-\sin\dfrac{4\pi}{7} \right)\\& = \dfrac{1}{2\sin\dfrac{2\pi}{7}}\left(\sin\dfrac{5\pi}{7}-2\cos\dfrac{7\pi}{2\cdot7}\cdot \sin\dfrac{\pi}{7}\right) \\& = \dfrac{\sin\dfrac{5\pi}{7}}{2\sin\dfrac{2\pi}{7}}= \dfrac{\sin\left(\pi-\dfrac{2\pi}{7}\right)}{2\sin\dfrac{2\pi}{7}} =\dfrac{1}{2}\left(\dfrac{\sin\dfrac{2\pi}{7}}{\sin\dfrac{2\pi}{7}}\right)=\dfrac{1}{2} \end{aligned}$