`(sec8A-1)/(sec4A-1)=?`

Options

A. `(tan2A)/(tan8A)`
B. `(tan8A)/(tan2A)`
C. `(cot8A)/(cot2A)`
D. `(cot2A)/(cot8A)`

Answer

To solve the expression \((\sec 8A - 1)/(\sec 4A - 1)\), we will follow these steps:

### Step 1: Rewrite secant in terms of cosine
Recall that \(\sec A = \frac{1}{\cos A}\). Therefore, we can rewrite the expression as:

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

Similarly, for \(\sec 4A - 1\):

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

### Step 2: Substitute these into the original expression
Now we substitute these into the original expression:

\[ \frac{\sec 8A - 1}{\sec 4A - 1} = \frac{\frac{1 - \cos 8A}{\cos 8A}}{\frac{1 - \cos 4A}{\cos 4A}} = \frac{(1 - \cos 8A) \cdot \cos 4A}{(1 - \cos 4A) \cdot \cos 8A} \]

### Step 3: Simplify the expression
Now we can simplify this expression further. We can use the identity \(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\):

\[ 1 - \cos 8A = 2 \sin^2(4A) \quad \text{and} \quad 1 - \cos 4A = 2 \sin^2(2A) \]

Substituting these into our expression gives:

\[ \frac{2 \sin^2(4A) \cdot \cos 4A}{2 \sin^2(2A) \cdot \cos 8A} \]

The 2's cancel out:

\[ \frac{\sin^2(4A) \cdot \cos 4A}{\sin^2(2A) \cdot \cos 8A} \]

### Step 4: Further simplification
Using the identity \(\sin(2x) = 2 \sin(x) \cos(x)\), we can express \(\sin^2(4A)\) as:

\[ \sin^2(4A) = \sin^2(2 \cdot 2A) = 4 \sin^2(2A) \cos^2(2A) \]

Substituting this back into the expression gives:

\[ \frac{4 \sin^2(2A) \cos^2(2A) \cdot \cos 4A}{\sin^2(2A) \cdot \cos 8A} \]

Now, we can cancel \(\sin^2(2A)\):

\[ \frac{4 \cos^2(2A) \cdot \cos 4A}{\cos 8A} \]

### Final Expression
Thus, the final expression simplifies to:

\[ \frac{4 \cos^2(2A) \cdot \cos 4A}{\cos 8A} \]

### Summary
The final result is:

\[ \frac{4 \cos^2(2A) \cdot \cos 4A}{\cos 8A} \]