`(1-tanx)/(1+tanx)=tanyandx-y=pi/6,t h e nx , y`

Options

A. `((5pi)/(24), (pi)/24)`,
B. `(-(7pi)/24, -(11pi)/(24))`
C. `(-(115pi)/(24), (-119pi)/(24))`
D. `((5pi)/24, (13pi)/24)`

Answer

To solve the problem \(\frac{1 - \tan x}{1 + \tan x} = \tan y\) and \(x - y = \frac{\pi}{6}\), we will follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ \frac{1 - \tan x}{1 + \tan x} = \tan y \]
We can use the tangent subtraction formula, which states that:
\[ \tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x} \]
So, we can rewrite the equation as:
\[ \tan\left(\frac{\pi}{4} - x\right) = \tan y \]

### Step 2: Set up the equation
From the equality of tangents, we can say:
\[ \frac{\pi}{4} - x = y + n\pi \quad (n \in \mathbb{Z}) \]
For simplicity, we will consider \(n = 0\):
\[ \frac{\pi}{4} - x = y \]
This can be rearranged to:
\[ x + y = \frac{\pi}{4} \quad \text{(Equation 1)} \]

### Step 3: Use the second equation
We are also given:
\[ x - y = \frac{\pi}{6} \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have a system of equations:
1. \(x + y = \frac{\pi}{4}\)
2. \(x - y = \frac{\pi}{6}\)

We can add these two equations:
\[ (x + y) + (x - y) = \frac{\pi}{4} + \frac{\pi}{6} \]
This simplifies to:
\[ 2x = \frac{\pi}{4} + \frac{\pi}{6} \]

### Step 5: Find a common denominator
To add \(\frac{\pi}{4}\) and \(\frac{\pi}{6}\), we need a common denominator:
The least common multiple of 4 and 6 is 12. Thus:
\[ \frac{\pi}{4} = \frac{3\pi}{12}, \quad \frac{\pi}{6} = \frac{2\pi}{12} \]
So,
\[ \frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi}{12} + \frac{2\pi}{12} = \frac{5\pi}{12} \]
Now we have:
\[ 2x = \frac{5\pi}{12} \]

### Step 6: Solve for \(x\)
Dividing both sides by 2 gives:
\[ x = \frac{5\pi}{24} \]

### Step 7: Substitute to find \(y\)
Now we substitute \(x\) back into Equation 1:
\[ \frac{5\pi}{24} + y = \frac{\pi}{4} \]
Substituting \(\frac{\pi}{4}\) with \(\frac{6\pi}{24}\):
\[ \frac{5\pi}{24} + y = \frac{6\pi}{24} \]
Subtracting \(\frac{5\pi}{24}\) from both sides:
\[ y = \frac{6\pi}{24} - \frac{5\pi}{24} = \frac{\pi}{24} \]

### Final Answer
Thus, the values of \(x\) and \(y\) are:
\[ x = \frac{5\pi}{24}, \quad y = \frac{\pi}{24} \]