If `f(x)=cos^(2)x+sec^(2)x`, then

Options

A. f(x) `lt`1
B. f(x) = 1
C. 2`lt`f(x)`lt`1
D. f(x)`ge`2

Explanation

To solve the problem \( f(x) = \cos^2 x + \sec^2 x \), we will use the concepts of Arithmetic Mean (AM) and Geometric Mean (GM).

### Step-by-Step Solution:

1. **Identify the Functions**:
We have the function:
\[ f(x) = \cos^2 x + \sec^2 x \]

2. **Use the AM-GM Inequality**:
According to the AM-GM inequality, for any two non-negative numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
Here, let \( a = \cos^2 x \) and \( b = \sec^2 x \).

3. **Apply AM-GM to \( f(x) \)**:
By applying the AM-GM inequality:
\[ \frac{\cos^2 x + \sec^2 x}{2} \geq \sqrt{\cos^2 x \cdot \sec^2 x} \]

4. **Simplify the Right Side**:
We know that:
\[ \sec x = \frac{1}{\cos x} \implies \sec^2 x = \frac{1}{\cos^2 x} \]
Therefore:
\[ \cos^2 x \cdot \sec^2 x = \cos^2 x \cdot \frac{1}{\cos^2 x} = 1 \]
Thus:
\[ \sqrt{\cos^2 x \cdot \sec^2 x} = \sqrt{1} = 1 \]

5. **Combine the Results**:
From the AM-GM inequality, we have:
\[ \frac{\cos^2 x + \sec^2 x}{2} \geq 1 \]
Multiplying both sides by 2 gives:
\[ \cos^2 x + \sec^2 x \geq 2 \]

6. **Conclusion**:
Therefore, we conclude that:
\[ f(x) \geq 2 \]

### Final Result:
The minimum value of \( f(x) = \cos^2 x + \sec^2 x \) is \( 2 \).