Find the range of `f(x)=(1)/(|sinx|)+(1)/(|cosx|)`

Answer

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

### Step 1: Identify the domain of the function
The function \( f(x) \) involves \( |\sin x| \) and \( |\cos x| \) in the denominators. Since both sine and cosine can take values between -1 and 1, their absolute values will range from 0 to 1. However, the function is undefined when \( |\sin x| = 0 \) or \( |\cos x| = 0 \). This occurs at:
- \( x = n\pi \) for \( |\sin x| = 0 \)
- \( x = \frac{\pi}{2} + n\pi \) for \( |\cos x| = 0 \)

Thus, the domain of \( f(x) \) excludes these points.

### Step 2: Analyze the behavior of the function
As \( |\sin x| \) and \( |\cos x| \) approach 0, \( f(x) \) approaches infinity. Conversely, when \( |\sin x| \) and \( |\cos x| \) approach 1 (which occurs at \( x = \frac{\pi}{4} + n\frac{\pi}{2} \)), \( f(x) \) will take its minimum value.

### Step 3: Calculate the minimum value of \( f(x) \)
To find the minimum value, we can use the fact that \( |\sin x| + |\cos x| \) is maximized when both are equal. This occurs at \( x = \frac{\pi}{4} + n\frac{\pi}{2} \):
\[ |\sin x| = |\cos x| = \frac{1}{\sqrt{2}} \]
Substituting this into \( f(x) \):
\[ f\left(\frac{\pi}{4}\right) = \frac{1}{\frac{1}{\sqrt{2}}} + \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \]

### Step 4: Determine the range of the function
Since \( f(x) \) approaches infinity as \( |\sin x| \) or \( |\cos x| \) approaches 0, and has a minimum value of \( 2\sqrt{2} \), the range of \( f(x) \) is:
\[ [2\sqrt{2}, \infty) \]

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