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The expression given in the problem is:

\[
\sec^2 \left( \frac{1}{4} \cos^{-1} \left( \frac{1}{8} \right) \right)
\]

Let's go step-by-step to solve this:

### Step 1: Find \( \cos^{-1} \left( \frac{1}{8} \right) \)
The inverse cosine function, \( \cos^{-1}(x) \), returns the angle whose cosine is \( x \). So we need to find an angle \( \theta \) such that:

\[
\cos(\theta) = \frac{1}{8}
\]

Let this angle be \( \theta = \cos^{-1} \left( \frac{1}{8} \right) \).

### Step 2: Multiply the result by \( \frac{1}{4} \)
The expression now becomes:

\[
\frac{1}{4} \times \theta
\]

### Step 3: Find \( \sec^2 \left( \frac{1}{4} \times \theta \right) \)
Recall that \( \sec(\theta) = \frac{1}{\cos(\theta)} \), so \( \sec^2(\theta) = \frac{1}{\cos^2(\theta)} \).

We need to determine the value of this secant expression based on the inverse cosine and proceed to find a possible value for it.

Let's calculate this value.The value of \( \sec^2 \left( \frac{1}{4} \cos^{-1} \left( \frac{1}{8} \right) \right) \) is approximately \( 1.14 \).

Comparing this with the given options:

1. \( \frac{7}{3} = 2.33 \)
2. \( \frac{8}{7} = 1.14 \) (this matches the calculated value)
3. \( \frac{5}{3} = 1.67 \)
4. \( \frac{3}{2} = 1.5 \)

Thus, the correct answer is \( \frac{8}{7} \), which corresponds to option (2).

Would you like further details or have any questions? Here are some related questions for further practice:

1. How do you derive \( \sec^2(\theta) \) from a cosine value?
2. What is the domain of the \( \cos^{-1}(x) \) function?
3. How does \( \sec^2(\theta) \) relate to the Pythagorean identity?
4. Can you generalize the approach for other values of \( x \) in \( \cos^{-1}(x) \)?
5. How would the result change if we used \( \sin^{-1} \) instead of \( \cos^{-1} \)?

**Tip:** Always check the domain and range of inverse trigonometric functions to ensure valid inputs!

A possible value of sec²(1/4 cos⁻¹(1/8)) is:

Learn how to find a possible value of sec²(1/4 cos⁻¹(1/8)) using trigonometric identities and the inverse cosine function. This problem explores inverse trigonometric functions and secant calculations, suitable for high school students in Grades 11-12 or higher.

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