For $0<\theta <\pi /4,\frac{\sec \left(8\theta \right)-1}{\sec \left(4\theta \right)-1}\cdot \frac{\tan \left(2\theta \right)}{\tan \left(8\theta \right)}$ is equal to 

$\frac{\sec \left(8\theta \right)-1}{\sec \left(4\theta \right)-1}\cdot \frac{\tan \left(2\theta \right)}{\tan \left(8\theta \right)}$

$\begin{array}{l}=\frac{1-\cos \left(8\theta \right)}{1-\cos \left(4\theta \right)}\cdot \frac{\cos \left(4\theta \right)}{\cos \left(8\theta \right)}\cdot \frac{\sin \left(2\theta \right)}{\cos \left(2\theta \right)}\cdot \frac{\cos \left(8\theta \right)}{\sin \left(8\theta \right)}\\ =\frac{2{\sin }^2\left(4\theta \right)}{2{\sin }^2\left(2\theta \right)}\frac{\cos \left(4\theta \right)\sin \left(2\theta \right)}{\cos \left(2\theta \right)\sin \left(8\theta \right)}=1\end{array}$