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 a systematic approach.

### Step 1: Identify the domain of the function
The function \( f(x) \) is defined as long as \( |\sin x| \) and \( |\cos x| \) are not equal to zero. This occurs when:
- \( |\sin x| = 0 \) at \( x = n\pi \) (where \( n \) is any integer)
- \( |\cos x| = 0 \) at \( x = \frac{(2n+1)\pi}{2} \) (where \( n \) is any integer)

Thus, the function is undefined at these points.

**Hint**: Check where the sine and cosine functions equal zero to determine the domain.

### Step 2: Analyze the behavior of the function
Since \( |\sin x| \) and \( |\cos x| \) both vary between 0 and 1, we can analyze the function by considering the values of \( |\sin x| \) and \( |\cos x| \) approaching these limits.

### Step 3: Use inequalities
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
\[ \frac{1}{|\sin x|} + \frac{1}{|\cos x|} \geq 2 \sqrt{\frac{1}{|\sin x| \cdot |\cos x|}} \]

### Step 4: Substitute using the identity
We know that:
\[ |\sin x \cdot \cos x| = \frac{1}{2} |\sin 2x| \]
Thus, we can rewrite the inequality:
\[ f(x) \geq 2 \sqrt{\frac{2}{|\sin 2x|}} \]

### Step 5: Find the minimum value
The maximum value of \( |\sin 2x| \) is 1. Therefore, the minimum value of \( f(x) \) occurs when \( |\sin 2x| = 1 \):
\[ f(x) \geq 2 \sqrt{2} \]

### Step 6: Determine the range
Since \( f(x) \) can take any value greater than or equal to \( 2\sqrt{2} \) and can approach infinity, we conclude:
\[ \text{Range of } f(x) = [2\sqrt{2}, \infty) \]

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

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