If `x+y=pi//4` and `tan x+tan y=1`, then `(n in Z)`

Options

A. `sin x=0` always
B. when `x=npi+pi//4` then `y=-npi`
C. when `x=npi` then `y=npi +(pi//4)`
D. when `x=npi+pi//4` then `y=npi -(pi//4)`

Explanation

To solve the problem, we need to find the values of \( n \in \mathbb{Z} \) given the equations \( x + y = \frac{\pi}{4} \) and \( \tan x + \tan y = 1 \).

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ x + y = \frac{\pi}{4} \]
\[ \tan x + \tan y = 1 \]

2. **Use the identity for tangent:**
We know that:
\[ \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]
Substituting \( x + y = \frac{\pi}{4} \):
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
Therefore:
\[ 1 = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]
Since \( \tan x + \tan y = 1 \), we can substitute:
\[ 1 = \frac{1}{1 - \tan x \tan y} \]

3. **Cross-multiply to find \( \tan x \tan y \):**
\[ 1 - \tan x \tan y = 1 \implies \tan x \tan y = 0 \]

4. **Analyze the product of tangents:**
Since \( \tan x \tan y = 0 \), either \( \tan x = 0 \) or \( \tan y = 0 \).

5. **Case 1: \( \tan x = 0 \)**
This implies:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]
Then from \( x + y = \frac{\pi}{4} \):
\[ n\pi + y = \frac{\pi}{4} \implies y = \frac{\pi}{4} - n\pi \]

6. **Case 2: \( \tan y = 0 \)**
This implies:
\[ y = m\pi \quad (m \in \mathbb{Z}) \]
Then from \( x + y = \frac{\pi}{4} \):
\[ x + m\pi = \frac{\pi}{4} \implies x = \frac{\pi}{4} - m\pi \]

7. **Combine results:**
From both cases, we have:
- If \( x = n\pi \), then \( y = \frac{\pi}{4} - n\pi \).
- If \( y = m\pi \), then \( x = \frac{\pi}{4} - m\pi \).

8. **Final values of \( x \) and \( y \):**
Thus, we can express:
\[ x = n\pi + \frac{\pi}{4} \quad \text{and} \quad y = \frac{\pi}{4} - n\pi \]
or
\[ y = m\pi + \frac{\pi}{4} \quad \text{and} \quad x = \frac{\pi}{4} - m\pi \]

9. **Conclusion:**
The values of \( n \) can be any integer, hence \( n \in \mathbb{Z} \).