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 where \( f(x) = \cos^2 x + \sec^2 x \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \cos^2 x + \sec^2 x \]
We know that \( \sec^2 x = \frac{1}{\cos^2 x} \). Therefore, we can rewrite \( f(x) \) as:
\[ f(x) = \cos^2 x + \frac{1}{\cos^2 x} \]

### Step 2: Set \( y = \cos^2 x \)
Let \( y = \cos^2 x \). Then, we can express \( f(x) \) in terms of \( y \):
\[ f(x) = y + \frac{1}{y} \]

### Step 3: Find the minimum value of \( f(x) \)
To find the minimum value of \( f(x) \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that for any positive numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
Applying this to our function:
\[ \frac{y + \frac{1}{y}}{2} \geq \sqrt{y \cdot \frac{1}{y}} = \sqrt{1} = 1 \]
Multiplying both sides by 2 gives:
\[ y + \frac{1}{y} \geq 2 \]
Thus, we have:
\[ f(x) \geq 2 \]

### Step 4: Conclusion
The minimum value of \( f(x) \) occurs when \( y = 1 \) (which happens when \( \cos^2 x = 1 \), or \( x = n\pi \) for any integer \( n \)). Therefore, we conclude that:
\[ f(x) \geq 2 \]

### Final Answer
The minimum value of \( f(x) \) is \( 2 \).

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