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