The least value of `cos^(2)x+sec^(2)x` is :

Options

A. 1
B. 0
C. `-1`
D. 2

Answer

To find the least value of \( \cos^2 x + \sec^2 x \), we can use the relationship between cosine and secant, along with the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Recall the Definition of Secant**:
\[ \sec^2 x = \frac{1}{\cos^2 x} \]
Therefore, we can rewrite the expression:
\[ \cos^2 x + \sec^2 x = \cos^2 x + \frac{1}{\cos^2 x} \]

2. **Let \( y = \cos^2 x \)**:
This means we can express our function as:
\[ f(y) = y + \frac{1}{y} \]
where \( y > 0 \) since \( \cos^2 x \) is always positive.

3. **Use the AM-GM Inequality**:
According to the AM-GM inequality:
\[ \frac{y + \frac{1}{y}}{2} \geq \sqrt{y \cdot \frac{1}{y}} = 1 \]
Multiplying both sides by 2 gives:
\[ y + \frac{1}{y} \geq 2 \]

4. **Conclusion**:
The minimum value of \( \cos^2 x + \sec^2 x \) occurs when:
\[ y + \frac{1}{y} = 2 \]
This happens when \( y = 1 \) (or \( \cos^2 x = 1 \)), which implies \( \cos x = 1 \) or \( x = 2n\pi \) for any integer \( n \).

Thus, the least value of \( \cos^2 x + \sec^2 x \) is:
\[ \boxed{2} \]