For `theta>pi/3`, the value of `f(theta)=sec^2theta+cos^2theta` always lies in the interval

Options

A. 0,2
B. 0,1
C. 1,2
D. `2,oo`

Answer

To solve the problem, we need to analyze the function \( f(\theta) = \sec^2 \theta + \cos^2 \theta \) for \( \theta > \frac{\pi}{3} \).

### Step 1: Understand the function
The function consists of two parts: \( \sec^2 \theta \) and \( \cos^2 \theta \). We know that:
- \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \)
- \( \cos^2 \theta \) is the square of the cosine function.

### Step 2: Analyze the range of \( \cos \theta \)
For \( \theta > \frac{\pi}{3} \):
- The value of \( \cos \theta \) decreases from \( \frac{1}{2} \) (at \( \theta = \frac{\pi}{3} \)) to \( 0 \) (as \( \theta \) approaches \( \frac{\pi}{2} \)).
- Therefore, \( \cos^2 \theta \) will range from \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \) to \( 0 \).

### Step 3: Analyze \( \sec^2 \theta \)
Since \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \):
- As \( \cos^2 \theta \) approaches \( 0 \), \( \sec^2 \theta \) approaches \( +\infty \).
- When \( \cos^2 \theta = \frac{1}{4} \), \( \sec^2 \theta = \frac{1}{\frac{1}{4}} = 4 \).

### Step 4: Combine the two parts
Now we can find the range of \( f(\theta) \):
- At \( \cos^2 \theta = \frac{1}{4} \):
\[ f(\theta) = 4 + \frac{1}{4} = 4.25 \]
- As \( \cos^2 \theta \) approaches \( 0 \):
\[ f(\theta) \to +\infty \]

### Step 5: Conclusion
Thus, the function \( f(\theta) \) for \( \theta > \frac{\pi}{3} \) lies in the interval:
\[ [4.25, +\infty) \]

### Final Answer
The value of \( f(\theta) = \sec^2 \theta + \cos^2 \theta \) always lies in the interval \([4.25, +\infty)\).

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