Minimum value of `sec ^(2) theta + cos ^(2) theta` is

Options

A. 1
B. 0
C. 2
D. none of these

Answer

To find the minimum value of \( \sec^2 \theta + \cos^2 \theta \), we can follow these steps:

### Step 1: Rewrite the expression
We know that \( \sec \theta = \frac{1}{\cos \theta} \). Therefore, we can express \( \sec^2 \theta \) in terms of \( \cos \theta \):
\[ \sec^2 \theta = \frac{1}{\cos^2 \theta} \]
Thus, we can rewrite the expression as:
\[ \sec^2 \theta + \cos^2 \theta = \frac{1}{\cos^2 \theta} + \cos^2 \theta \]

### Step 2: Let \( x = \cos^2 \theta \)
Let \( x = \cos^2 \theta \). Then, the expression becomes:
\[ \frac{1}{x} + x \]
where \( 0 < x \leq 1 \) since \( \cos^2 \theta \) ranges between 0 and 1.

### Step 3: Find the derivative
To find the minimum value, we can differentiate the function \( f(x) = \frac{1}{x} + x \) with respect to \( x \):
\[ f'(x) = -\frac{1}{x^2} + 1 \]
Setting the derivative to zero to find critical points:
\[ -\frac{1}{x^2} + 1 = 0 \implies \frac{1}{x^2} = 1 \implies x^2 = 1 \implies x = 1 \]

### Step 4: Evaluate the second derivative
To confirm that this critical point is a minimum, we can evaluate the second derivative:
\[ f''(x) = \frac{2}{x^3} \]
Since \( f''(x) > 0 \) for \( x > 0 \), this indicates that \( x = 1 \) is indeed a minimum.

### Step 5: Calculate the minimum value
Now, substituting \( x = 1 \) back into the expression:
\[ f(1) = \frac{1}{1} + 1 = 1 + 1 = 2 \]

### Conclusion
Thus, the minimum value of \( \sec^2 \theta + \cos^2 \theta \) is:
\[ \boxed{2} \]