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?

As $\cos2y=1-2\sin^2y,\sin2y=2\sin y\cos y$
$$\dfrac{1-\sec16A}{\tan16A}=\cdots=\dfrac{\cos16A-1}{\sin16A}=-\dfrac{2\sin^28A}{2\cos8A\sin8A}=-\tan8A$$
Similarly, $$\dfrac{1-\sec8A}{\tan8A}=-\tan4A$$
Can you take it home from here?
