Prove that $\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$

Question

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?

Answer 1

$\frac{\sec 16A -1}{\sec 8A -1}$

= $\frac{\frac{1}{\cos 16A}-1}{\frac{1}{\cos 8A}-1}$

= $\frac{\frac{1 - \cos 16A}{\cos 16A}}{\frac{1 - \cos 8A}{\cos 8A}}$

= $\frac{2 \sin^2 8A}{\cos 16A} × \frac{\cos 8A}{2 \sin^2 4A}$

= $\frac{2 \sin 8A \cos 8A}{\cos 16A} × \frac{\sin 8A}{2 \sin^2 4A}$

= $\frac{\sin 16A}{\cos 16A} × \frac{ 2 \sin 4A \cos 4A}{2 \sin^2 4A}$

= $\tan 16A × \frac{\cos 4A}{\sin 4A}$

= $\tan 16A × \cot 4A$

= $\frac{\tan 16A}{\tan 4A}$

Answer 2

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

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

Answer 3

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?