Question

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

Answer 1

$\frac{\sec 8\theta -1}{\sec 4\theta -1}$

$=\frac{\frac{\tan 8\theta }{\tan 8\theta }(\sec 8\theta -1)}{\frac{\tan 4\theta }{\tan 4\theta }(\sec 4\theta -1)}$

$=\frac{\tan 8\theta \left(\frac{1-\cos 8\theta }{\sin 8\theta }\right)}{\tan 4\theta \left(\frac{1-\cos 4\theta }{\sin 4\theta }\right)}$

$=\frac{\tan 8\theta \times \tan 4\theta }{\tan 4\theta \times \tan 2\theta }$

$\therefore \frac{\sec 8\theta -1}{\sec 4\theta -1}=\frac{\tan 8\theta }{\tan 2\theta }$