Prove that $\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$ Prove that:$$\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$$.
My Attempt,
$$L.H.S= \frac {\sec (16A)-1}{\sec (8A)-1}$$
$$=\frac {\frac {1}{\cos (16A)} -1}{\frac {1}{\cos (8A)} -1}$$
$$=\frac {(1-\cos (16A)).(\cos (8A)}{(\cos (16A))(1-\cos (8A))}$$.
What should I do next?

$\frac{\sec 16A -1}{\sec 8A -1}$
= $\frac{\frac{1}{\cos 16A}-1}{\frac{1}{\cos 8A}-1}$
= $\frac{\frac{1 - \cos 16A}{\cos 16A}}{\frac{1 - \cos 8A}{\cos 8A}}$
= $\frac{2 \sin^2 8A}{\cos 16A} × \frac{\cos 8A}{2 \sin^2 4A}$
= $\frac{2 \sin 8A \cos 8A}{\cos 16A} × \frac{\sin 8A}{2 \sin^2 4A}$
= $\frac{\sin 16A}{\cos 16A} × \frac{ 2 \sin 4A \cos 4A}{2 \sin^2 4A}$
= $\tan 16A × \frac{\cos 4A}{\sin 4A}$
= $\tan 16A × \cot 4A$
= $\frac{\tan 16A}{\tan 4A}$
