If `f(x)=cos^2x+s e c^2x ,t h e n` `(a)f(x)<1` (b) `f(x)=1` (c) `2ltf(x)<1` (d) `""""f(x)ge2`

Explanation

To solve the problem, we need to analyze the function \( f(x) = \cos^2 x + \sec^2 x \).

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

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