Prove that $\frac{\sec 8A-1}{\sec 4A-1}=\frac{\tan 8A}{\tan 2A}$ 
Prove that $$\frac{\sec 8A-1}{\sec 4A-1}=\frac{\tan 8A}{\tan 2A}$$

I tried converting $\sec 4A$ into $\dfrac{1}{\cos 4A}$ and $\sec 8A$ into $\dfrac{1}{\cos 8A}$ but couldn't get the answer. Any help would be appreciated.
 A: Using $\cos2x=1-2\sin^2x,\sin2x=2\sin x\cos x$
$$\dfrac{\sec2x-1}{\tan2x}=\dfrac{1-\cos2x}{\sin2x}=\dfrac{2\sin^2x}{2\sin x\cos x}=\tan x$$
$$\implies\sec2x-1=\tan2x\tan x$$
Set $2x=8A,4A$ and divide.
A: Converting $\sec \to \cos$,
$$(\frac{1-\cos (8A)}{1-\cos (4A)})(\frac{\cos( 4A)}{\cos (8A)})$$
$$(\frac{2\sin^2 (4A)}{1-\cos (4A)})(\frac{\cos (4A)}{\cos (8A)})$$
$$(\frac{\sin (8A)×\sin (4A)}{1-\cos (4A)})(\frac{1}{\cos (8A)})$$
$$\frac{\tan (8A)×\sin (4A)}{1-\cos (4A)}$$
$$\frac{2\tan (8A)×\sin( 2A)×\cos (2A)}{2\sin^2 (2A)}$$
$$\frac{\tan (8A)×\cos (2A)}{\sin (2A)}$$
$$\frac{\tan (8A)}{\tan( 2A)}$$
A: $$\quad{\frac{\sec 8a-1}{\sec 4a-1}=\frac{\frac{1}{\cos 8a}-1}{\frac{1}{\cos 4a}-1}=\\
\frac{\cos 4a}{\cos 8a}.\frac{1-\cos 8a}{1-\cos 4a}=\\
\frac{\cos 4a}{\cos 8a}.\frac{2\sin^24a}{2\sin^22a}=\\
\frac{\cos 4a.2\sin 4a}{\cos 8a}.\frac{\sin 4a}{2\sin^22a}=\\
\frac{\sin 8a}{\cos 8a}.\frac{\sin 4a}{2\sin^22a}=\\
\frac{\sin 8a}{\cos 8a}.\frac{2\sin 2a.\cos 2a}{2\sin^22a}=\\
\frac{\sin 8a}{\cos 8a}.\frac{\cos 2a}{\sin 2a}=\\
\underbrace{\frac{\sin 8a}{\cos 8a}}_{\tan 8a}.\underbrace{\frac{\cos 2a}{\sin 2a}}_{\cot 2a}=\\}$$
A: $$\frac{\sec8a-1}{\sec4a-1}=\frac{\cos4a(1-\cos8a)}{\cos8a(1-\cos4a)}=\frac{\cos4a\sin^24a}{\cos8a\sin^22a}=$$
$$=\frac{2\cos4a\sin4a\sin2a\cos2a}{\cos8a\sin^22a}=\frac{\sin8a\cos2a}{\cos8a\sin2a}=\frac{\tan8a}{\tan2a}.$$
A: $\DeclareMathOperator{exsec}{\mathrm{exsec}}$You can solve this by exploiting the identity (50).
$$ \tan{t}\tan{2t} = \exsec{t} \tag{50} $$
First a bit of notation, let $x(t) = \cos t$ and $y(t) = \sin(t)$ . Let $x$ and $y$ refer to $x(t)$ and $y(t)$ as an abbreviation.
Let's prove (50).
$$ \tan{2t} \times \tan t \tag{101} $$
$$ \frac{y(2t)}{x(2t)} \times \tan t \tag{102} $$
$$ \frac{2xy}{x^2 - y^2} \times \tan t \tag{103} $$
$$ \frac{2xy}{x^2 - y^2} \times \frac{y}{x} \tag{104} $$
$$ \frac{2y^2}{x^2 - y^2} \tag{105} $$
$$ \frac{1 - x^2 + y^2}{x^2 - y^2} \tag{106} $$
$$ \frac{1}{x^2-y^2} - \frac{(-x^2) + y^2}{x^2 - y^2} \tag{107} $$
$$ \frac{1}{x(2t)} - 1 \tag{108} $$
$$ \sec(2t) - 1 \tag{109} $$
$$ \exsec(2t) \tag{110} $$
Now that we've proven (50), let's replace $t$ with $4a$ (201).
$$ \tan(8a) \times \tan(4a) = \exsec(8a) \tag{201} $$
Let's replace $t$ with $2a$ (202).
$$ \tan(4a) \times \tan(2a) = \exsec(4a) \tag{202} $$
Let's take the ratio of (201) and (202) yielding (203), which simplifies to (204).
$$ \frac{\tan(8a) \times \tan(4a)}{\tan(4a) \times \tan(2a)} = \frac{\exsec{8a}}{\exsec{4a}} \tag{203} $$
$$ \frac{\tan(8a)}{\tan(2a)} = \frac{\exsec(8a)}{\exsec(4a)} \tag{204} $$
(204) is straightforwardly equivalent to (205), what we wanted to show.
$$ \frac{\sec(8A) - 1}{\sec(4A) - 1} = \frac{\tan(8A)}{\tan(2A)} \tag{205} $$
