The range of the function
`f(x)=1/abs(sinx)+1/abs(cosx)` is

Options

A. `[2sqrt(2),infty)`
B. `(sqrt(2),2sqrt(2))`
C. `(0,2sqrt(2))`
D. `(2sqrt(2),4)`

Answer

To find the range of the function \( f(x) = \frac{1}{|\sin x|} + \frac{1}{|\cos x|} \), we can follow these steps:

### Step 1: Identify the domain of the function
Since \( f(x) \) involves \( |\sin x| \) and \( |\cos x| \) in the denominator, we need to ensure that these values are not zero. Therefore, we have:
- \( \sin x \neq 0 \) which implies \( x \neq n\pi \) for any integer \( n \).
- \( \cos x \neq 0 \) which implies \( x \neq \frac{(2n + 1)\pi}{2} \) for any integer \( n \).

### Step 2: Analyze the behavior of \( |\sin x| \) and \( |\cos x| \)
The values of \( |\sin x| \) and \( |\cos x| \) both range from 0 to 1. However, they cannot be equal to 0, so:
- \( 0 < |\sin x| \leq 1 \)
- \( 0 < |\cos x| \leq 1 \)

### Step 3: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM)
Using the AM-GM inequality:
\[ \frac{1}{|\sin x|} + \frac{1}{|\cos x|} \geq 2 \sqrt{\frac{1}{|\sin x| \cdot |\cos x|}} \]
This implies:
\[ f(x) \geq 2 \sqrt{\frac{1}{|\sin x| \cdot |\cos x|}} \]

### Step 4: Express \( |\sin x| \cdot |\cos x| \)
Using the identity \( |\sin 2x| = 2 |\sin x| |\cos x| \), we can rewrite:
\[ |\sin x| \cdot |\cos x| = \frac{|\sin 2x|}{2} \]
Thus, we have:
\[ f(x) \geq 2 \sqrt{\frac{2}{|\sin 2x|}} \]

### Step 5: Determine the minimum value of \( f(x) \)
The maximum value of \( |\sin 2x| \) is 1, which occurs when \( 2x = \frac{\pi}{2} + n\pi \) for any integer \( n \). Therefore:
\[ f(x) \geq 2 \sqrt{2} \]

### Step 6: Conclusion about the range
Since \( f(x) \) can increase without bound as \( |\sin x| \) or \( |\cos x| \) approaches 0, the range of \( f(x) \) is:
\[ [2\sqrt{2}, \infty) \]

### Final Answer
The range of the function \( f(x) = \frac{1}{|\sin x|} + \frac{1}{|\cos x|} \) is \( [2\sqrt{2}, \infty) \).