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

Explanation

To find the domain of the function \( f(x) = \sqrt{\tan x - \tan^2 x} \), we need to ensure that the expression under the square root is non-negative. Here’s how we can solve it step by step:

### Step 1: Set the Inequality
We start by setting the expression under the square root greater than or equal to zero:
\[ \tan x - \tan^2 x \geq 0 \]

### Step 2: Factor the Expression
We can factor the left-hand side:
\[ \tan x (1 - \tan x) \geq 0 \]
This means we need to find where the product of \( \tan x \) and \( (1 - \tan x) \) is non-negative.

### Step 3: Identify Critical Points
The critical points occur when either factor is zero:
1. \( \tan x = 0 \) which gives \( x = n\pi \) for \( n \in \mathbb{Z} \).
2. \( 1 - \tan x = 0 \) which gives \( \tan x = 1 \) or \( x = \frac{\pi}{4} + n\pi \) for \( n \in \mathbb{Z} \).

### Step 4: Test Intervals
We will test the intervals determined by these critical points:
- The intervals are:
- \( (-\infty, n\pi) \)
- \( (n\pi, \frac{\pi}{4} + n\pi) \)
- \( (\frac{\pi}{4} + n\pi, (n+1)\pi) \)

We need to check the sign of \( \tan x (1 - \tan x) \) in each interval.

1. **Interval \( (n\pi, \frac{\pi}{4} + n\pi) \)**:
- Here, \( \tan x > 0 \) and \( 1 - \tan x > 0 \) (since \( \tan x < 1 \)).
- Thus, \( \tan x (1 - \tan x) > 0 \).

2. **Interval \( (\frac{\pi}{4} + n\pi, (n+1)\pi) \)**:
- Here, \( \tan x > 0 \) and \( 1 - \tan x < 0 \) (since \( \tan x > 1 \)).
- Thus, \( \tan x (1 - \tan x) < 0 \).

### Step 5: Combine Results
The expression \( \tan x (1 - \tan x) \geq 0 \) is satisfied in the intervals:
\[ [n\pi, \frac{\pi}{4} + n\pi] \]
where \( n \in \mathbb{Z} \).

### Conclusion
The domain of the function \( f(x) \) is:
\[ x \in [n\pi, n\pi + \frac{\pi}{4}] \quad \text{for } n \in \mathbb{Z} \]