`f(x)=sqrt(1-sin^(2)x)+sqrt(1+tan^(2)x)` then

Options

A. fundamental period of `f(x) " is " pi`
B. range of `f(x)" is "[2, oo)`
C. domain of `f(x)` is R
D. `f(x)=2` has 3 solution in `[0, 2pi]`

Explanation

To solve the problem given by the function \( f(x) = \sqrt{1 - \sin^2 x} + \sqrt{1 + \tan^2 x} \), we will simplify and analyze the function step by step.

### Step 1: Simplify \( f(x) \)
We start with the expression:
\[ f(x) = \sqrt{1 - \sin^2 x} + \sqrt{1 + \tan^2 x} \]

Using the Pythagorean identity, we know that:
\[ 1 - \sin^2 x = \cos^2 x \]
Thus, we can rewrite the first term:
\[ \sqrt{1 - \sin^2 x} = \sqrt{\cos^2 x} = |\cos x| \]

Next, we simplify the second term. Using the identity \( 1 + \tan^2 x = \sec^2 x \), we have:
\[ \sqrt{1 + \tan^2 x} = \sqrt{\sec^2 x} = |\sec x| \]

So, we can rewrite \( f(x) \) as:
\[ f(x) = |\cos x| + |\sec x| \]

### Step 2: Analyze the function
The function \( f(x) \) can be expressed as:
\[ f(x) = |\cos x| + \frac{1}{|\cos x|} \]
for \( \cos x \neq 0 \).

### Step 3: Determine the range of \( f(x) \)
To find the minimum value of \( f(x) \), we can use the AM-GM inequality:
\[ |\cos x| + \frac{1}{|\cos x|} \geq 2 \]
This inequality holds true for all positive values of \( |\cos x| \). The equality occurs when \( |\cos x| = 1 \), which happens when \( x = n\pi \) for \( n \in \mathbb{Z} \).

### Step 4: Domain of \( f(x) \)
The function \( f(x) \) is defined for all \( x \) except where \( \cos x = 0 \). This occurs at:
\[ x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]
Thus, the domain of \( f(x) \) is:
\[ \text{Domain} = \mathbb{R} \setminus \left\{ \frac{\pi}{2} + n\pi \, | \, n \in \mathbb{Z} \right\} \]

### Step 5: Conclusion
From the analysis:
- The minimum value of \( f(x) \) is 2.
- The function is defined for all real numbers except where \( \cos x = 0 \).

### Final Answer
Thus, we conclude:
1. The minimum value of \( f(x) \) is 2.
2. The domain of \( f(x) \) is \( \mathbb{R} \setminus \left\{ \frac{\pi}{2} + n\pi \, | \, n \in \mathbb{Z} \right\} \).