Prove that `(sec 8theta-1)/(sec 4 theta -1) = (tan 8theta)/(tan 2theta)`

Explanation

To prove that \(\frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{\tan 8\theta}{\tan 2\theta}\), we will start with the left-hand side (LHS) and manipulate it step by step until we reach the right-hand side (RHS).

### Step 1: Rewrite Secant in Terms of Cosine
We know that \(\sec \theta = \frac{1}{\cos \theta}\). Thus, we can rewrite the LHS as:
\[ \text{LHS} = \frac{\frac{1}{\cos 8\theta} - 1}{\frac{1}{\cos 4\theta} - 1} \]

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