The smallest positive values of x and y which satisfy `"tan" (x-y) =1, "sec" (x+y) = (2)/(sqrt(3))`, are

Options

A. `x = (25pi)/(24), y = (19pi)/(24)`
B. `x = (7pi)/(24), y = (37pi)/(24)`
C. `x = (pi)/(4), y = (pi)/(2)`
D. `x = (pi)/(3), y = (7pi)/(12)`

Answer

To solve the problem, we need to find the smallest positive values of \( x \) and \( y \) that satisfy the equations:

1. \( \tan(x - y) = 1 \)
2. \( \sec(x + y) = \frac{2}{\sqrt{3}} \)

### Step 1: Solve the first equation

From the first equation, we have:
\[ \tan(x - y) = 1 \]
We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \). Therefore, we can write:
\[ x - y = \frac{\pi}{4} + n\pi \quad \text{(where \( n \) is any integer)} \]
For the smallest positive solution, we can consider \( n = 0 \):
\[ x - y = \frac{\pi}{4} \quad \text{(Equation 1)} \]

### Step 2: Solve the second equation

From the second equation, we have:
\[ \sec(x + y) = \frac{2}{\sqrt{3}} \]
We know that \( \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} \). Thus, we can write:
\[ x + y = \frac{\pi}{6} + m \cdot 2\pi \quad \text{(where \( m \) is any integer)} \]
For the smallest positive solution, we can consider \( m = 0 \):
\[ x + y = \frac{\pi}{6} \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations

Now we have the 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:
\[ 2x = \frac{\pi}{4} + \frac{\pi}{6} \]
To add the fractions, we find a common denominator (which is 12):
\[ \frac{\pi}{4} = \frac{3\pi}{12}, \quad \frac{\pi}{6} = \frac{2\pi}{12} \]
So,
\[ 2x = \frac{3\pi}{12} + \frac{2\pi}{12} = \frac{5\pi}{12} \]
Thus,
\[ x = \frac{5\pi}{24} \]

### Step 4: Find \( y \)

Now we can substitute \( x \) back into one of the original equations to find \( y \). Using Equation 1:
\[ \frac{5\pi}{24} - y = \frac{\pi}{4} \]
Substituting \( \frac{\pi}{4} = \frac{6\pi}{24} \):
\[ \frac{5\pi}{24} - y = \frac{6\pi}{24} \]
Rearranging gives:
\[ -y = \frac{6\pi}{24} - \frac{5\pi}{24} = \frac{\pi}{24} \]
Thus,
\[ y = -\frac{\pi}{24} \]
Since we need the smallest positive values, we will consider the periodic nature of the functions involved.

### Step 5: Adjust for positive values

Since \( y \) is negative, we need to adjust our equations. We can consider the periodicity of the tangent and secant functions.

For secant, we can also consider:
\[ x + y = \frac{11\pi}{6} \quad \text{(as we need to find another solution)} \]
Now substituting \( x = \frac{5\pi}{24} \):
\[ \frac{5\pi}{24} + y = \frac{11\pi}{6} \]
Converting \( \frac{11\pi}{6} \) to a common denominator:
\[ \frac{11\pi}{6} = \frac{44\pi}{24} \]
So:
\[ y = \frac{44\pi}{24} - \frac{5\pi}{24} = \frac{39\pi}{24} \]

### Final Values

Thus, the smallest positive values of \( x \) and \( y \) that satisfy the equations are:
\[ x = \frac{5\pi}{24}, \quad y = \frac{39\pi}{24} \]