Sometimes functions are defined like `f(x)=max{sinx,cosx}`, then `f(x)` is splitted like `f(x)={{:(cosx, x in (0,(pi)/(4)]),(sinx, x in ((pi)/(4),(pi)/(2)]):}` etc.
If `f(x)=min{tanx, cotx}` then `f(x)=1` when `x=`

Options

A. `(npi)/(2)+(pi)/(4)`
B. `npi+(pi)/(6)`
C. `npi+(pi)/(4)`
D. `2npi+(pi)/(4)`

Answer

To solve the problem where \( f(x) = \min\{\tan x, \cot x\} \) and we need to find the values of \( x \) for which \( f(x) = 1 \), we can follow these steps:

### Step 1: Understand the Functions
We know that:
- \( \tan x = \frac{\sin x}{\cos x} \)
- \( \cot x = \frac{\cos x}{\sin x} \)

The function \( f(x) \) takes the minimum of these two values.

### Step 2: Set Up the Equation
To find when \( f(x) = 1 \), we need to solve the equations:
1. \( \tan x = 1 \)
2. \( \cot x = 1 \)

### Step 3: Solve \( \tan x = 1 \)
The equation \( \tan x = 1 \) holds true at:
\[ x = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 4: Solve \( \cot x = 1 \)
The equation \( \cot x = 1 \) also holds true at:
\[ x = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 5: Combine the Results
Since both equations yield the same solutions, we conclude that:
\[ f(x) = 1 \text{ when } x = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Final Answer
Thus, the values of \( x \) for which \( f(x) = 1 \) are:
\[ x = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

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