Find the domain of the function
`f (x) = sqrt(tan x - tan^(2) x)`

Answer

To find the domain of the function \( f(x) = \sqrt{\tan x - \tan^2 x} \), we need to ensure that the expression inside the square root is non-negative. This leads us to the inequality:

\[ \tan x - \tan^2 x \geq 0 \]

### Step 1: Factor the expression
We can factor the expression as follows:

\[ \tan x (1 - \tan x) \geq 0 \]

### Step 2: Analyze the factors
The product \( \tan x (1 - \tan x) \) is non-negative when both factors are either positive or zero, or both are negative.

1. **Case 1:** \( \tan x \geq 0 \) and \( 1 - \tan x \geq 0 \)
- This implies:
\[ \tan x \geq 0 \quad \text{and} \quad \tan x \leq 1 \]
- Therefore, we have:
\[ 0 \leq \tan x \leq 1 \]

2. **Case 2:** \( \tan x \leq 0 \) and \( 1 - \tan x \leq 0 \)
- This implies:
\[ \tan x \leq 0 \quad \text{and} \quad \tan x \geq 1 \]
- However, this case is impossible because \( \tan x \) cannot be both less than or equal to 0 and greater than or equal to 1 at the same time.

### Step 3: Determine the intervals for \( x \)
Now we need to find the intervals where \( 0 \leq \tan x \leq 1 \).

- The function \( \tan x \) is zero at \( x = n\pi \) (where \( n \) is any integer).
- The function \( \tan x \) equals 1 at \( x = \frac{\pi}{4} + n\pi \).

Thus, we need the intervals where \( \tan x \) is between 0 and 1. The relevant intervals are:

1. From \( n\pi \) to \( \frac{\pi}{4} + n\pi \) for \( n \) being any integer.

### Step 4: Write the domain
The domain of the function \( f(x) \) can be expressed as:

\[ \text{Domain of } f(x) = \bigcup_{n \in \mathbb{Z}} \left[ n\pi, \frac{\pi}{4} + n\pi \right] \]

### Final Answer
The domain of the function \( f(x) = \sqrt{\tan x - \tan^2 x} \) is:

\[ \boxed{\bigcup_{n \in \mathbb{Z}} \left[ n\pi, \frac{\pi}{4} + n\pi \right]} \]