$\frac{\sec 8 \theta-1}{\sec 4 \theta-1}=\frac{\frac{1}{\cos 8 \theta}-1}{\frac{1}{\cos 4 \theta}-1}=\frac{1-\cos 8 \theta}{\cos 8 \theta} \times \frac{\cos 4 \theta}{1-\cos 4 \theta}$

$=\frac{2 \sin ^{2} 4 \theta}{\cos 8 \theta} \times \frac{\cos 4 \theta}{2 \sin ^{2} 2 \theta} \quad\left[\because 1-\cos 8 \theta=2 \sin ^{2} \frac{8 \theta}{2}=2 \sin ^{2} 4 \theta\right]$ and

$\left[\because 1-\cos 4 \theta=2 \sin ^{2} \frac{4 \theta}{2}=2 \sin ^{2} 2 \theta\right]$

$=\frac{(2 \sin 4 \theta \cos 4 \theta)}{\cos 8 \theta} \times \frac{\sin 4 \theta}{2 \sin ^{2} 2 \theta}$

$=\left(\frac{2 \sin 4 \theta \cos 4 \theta}{\cos 8 \theta}\right) \times\left(\frac{2 \sin 2 \theta \cos 2 \theta}{2 \sin ^{2} 2 \theta}\right)$

$=\left(\frac{\sin 2(4 \theta)}{\cos 8 \theta}\right) \times\left(\frac{\cos 2 \theta}{\sin 2 \theta}\right)=\left(\frac{\sin 8 \theta}{\cos 8 \theta}\right) \times\left(\frac{\cos 2 \theta}{\sin 2 \theta}\right)$

$\tan 8 \theta \cot 2 \theta$