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.
[Math] Prove that $\frac{\sec 8A-1}{\sec 4A-1}=\frac{\tan 8A}{\tan 2A}$
trigonometry
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Best Answer
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.