Prove that \( \frac{\sec 8 \theta-1}{\sec 4 \theta-1}=\frac{\tan 8 \theta}{\tan 2 \theta} \)

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Prove that \( \frac{\sec 8 \theta-1}{\sec 4 \theta-1}=\frac{\tan 8 \theta}{\tan 2 \theta} \)
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