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?

$\displaystyle \frac{1-\cos^2(16 A)}{1+\cos (16A)}\times \frac{\cos 8 A}{\cos (16 A)} \times \frac{1+\cos 8A}{1-\cos^2 (8A)} = \frac{\sin^2 16A}{2\cos^2(8A)} \times \frac{\cos 8A}{\cos 16A} \times \frac{2\cos^2 4A}{\sin^2 8A}$
So $$ = \frac{\sin^2 (16 A)}{\sin (16A)} \times \frac{2\cos^2 (4A)}{\cos 16 A}\times \frac{1}{\sin (8A)} =  \frac{\tan 16 A}{\tan 4A}$$
