Prove that:`(sec8theta-1)/(sec4theta-1)=(tan8theta)/(tan2theta)`

Explanation

To prove that \(\frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{\tan 8\theta}{\tan 2\theta}\), we will start by rewriting the secant functions in terms of cosine and then simplify the expression step by step.

### Step 1: Rewrite secant in terms of cosine
We know that \(\sec x = \frac{1}{\cos x}\). Therefore, we can rewrite the left-hand side:

\[ \frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{\frac{1}{\cos 8\theta} - 1}{\frac{1}{\cos 4\theta} - 1} \]
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