the period of ` f(x) = 1/2 {(|sinx|)/(cos x)+(|cos x|)/(sin x)}` is

Options

A. ` (pi)/(2)`
B. ` pi`
C. ` 2 pi`
D. None

Answer

To find the period of the function \( f(x) = \frac{1}{2} \left( \frac{|\sin x|}{\cos x} + \frac{|\cos x|}{\sin x} \right) \), we need to determine the smallest positive value \( T \) such that \( f(x + T) = f(x) \) for all \( x \).

### Step 1: Check the period for \( T = \pi \)

Let's first check if \( T = \pi \) is a period of the function.

1. Compute \( f(x + \pi) \):
\[ f(x + \pi) = \frac{1}{2} \left( \frac{|\sin(x + \pi)|}{\cos(x + \pi)} + \frac{|\cos(x + \pi)|}{\sin(x + \pi)} \right) \]

Using the properties of sine and cosine:
- \( \sin(x + \pi) = -\sin x \)
- \( \cos(x + \pi) = -\cos x \)

Therefore:
\[ f(x + \pi) = \frac{1}{2} \left( \frac{|\sin x|}{-\cos x} + \frac{|\cos x|}{-\sin x} \right) = \frac{1}{2} \left( -\frac{|\sin x|}{\cos x} - \frac{|\cos x|}{\sin x} \right) \]

Since the absolute values make the terms positive:
\[ f(x + \pi) = -\frac{1}{2} \left( \frac{|\sin x|}{\cos x} + \frac{|\cos x|}{\sin x} \right) = -f(x) \]

Thus, \( f(x + \pi) \neq f(x) \). Hence, \( T = \pi \) is not a period.

### Step 2: Check the period for \( T = 2\pi \)

Now, let's check if \( T = 2\pi \) is a period of the function.

1. Compute \( f(x + 2\pi) \):
\[ f(x + 2\pi) = \frac{1}{2} \left( \frac{|\sin(x + 2\pi)|}{\cos(x + 2\pi)} + \frac{|\cos(x + 2\pi)|}{\sin(x + 2\pi)} \right) \]

Using the properties of sine and cosine:
- \( \sin(x + 2\pi) = \sin x \)
- \( \cos(x + 2\pi) = \cos x \)

Therefore:
\[ f(x + 2\pi) = \frac{1}{2} \left( \frac{|\sin x|}{\cos x} + \frac{|\cos x|}{\sin x} \right) = f(x) \]

Thus, \( f(x + 2\pi) = f(x) \). Hence, \( T = 2\pi \) is a period of the function.

### Conclusion

The period of the function \( f(x) = \frac{1}{2} \left( \frac{|\sin x|}{\cos x} + \frac{|\cos x|}{\sin x} \right) \) is \( 2\pi \).

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