If $\cos x = \frac{2 \cos y - 1}{2 - \cos y}$ where $x, y \in (0, \pi)$,then $\tan(x/2) \cot(y/2) =$

$3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right] = $

$\tan \frac{2 \pi}{7} \cdot \tan \frac{4 \pi}{7} + \tan \frac{4 \pi}{7} \cdot \tan \frac{\pi}{7} + \tan \frac{\pi}{7} \cdot \tan \frac{2 \pi}{7} = $

If $\sinh x = \tan A$,then $|\tanh x| =$

Evaluate the product: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$

If $x: y: z = \tan \left(\frac{\pi}{15}+\alpha\right): \tan \left(\frac{\pi}{15}+\beta\right): \tan \left(\frac{\pi}{15}+\gamma\right)$,then find the value of $\frac{z+x}{z-x} \sin ^2(\gamma-\alpha)+\frac{x+y}{x-y} \sin ^2(\alpha-\beta)+\frac{y+z}{y-z} \sin ^2(\beta-\gamma)$.