If `tan(alpha-beta)=1, sec(alpha+beta)=2/(sqrt(3)` and `alpha, beta` are positive then the smallest value of `alpha` is

Options

A. `142 (1^(@))/2`
B. `187 (1^(@))/2`
C. `7 (1^(@))/2`
D. `37 (1^(@))/2`

Answer

To solve the problem, we need to find the smallest value of \( \alpha \) given the equations \( \tan(\alpha - \beta) = 1 \) and \( \sec(\alpha + \beta) = \frac{2}{\sqrt{3}} \).

### Step 1: Solve for \( \alpha - \beta \)

Given that \( \tan(\alpha - \beta) = 1 \), we know that:

\[ \alpha - \beta = 45^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]

Since \( \alpha \) and \( \beta \) are positive angles, we can take \( n = 0 \):

\[ \alpha - \beta = 45^\circ \]

### Step 2: Solve for \( \alpha + \beta \)

Next, we have \( \sec(\alpha + \beta) = \frac{2}{\sqrt{3}} \). We know that:

\[ \sec(x) = \frac{1}{\cos(x)} \]

Thus, we can rewrite the equation as:

\[ \cos(\alpha + \beta) = \frac{\sqrt{3}}{2} \]

This implies:

\[ \alpha + \beta = 30^\circ + m \cdot 360^\circ \quad (m \in \mathbb{Z}) \]

Again, since \( \alpha \) and \( \beta \) are positive angles, we can take \( m = 0 \):

\[ \alpha + \beta = 30^\circ \]

### Step 3: Set up the system of equations

Now we have a system of equations:

1. \( \alpha - \beta = 45^\circ \)
2. \( \alpha + \beta = 30^\circ \)

### Step 4: Solve for \( \alpha \) and \( \beta \)

We can add these two equations:

\[ (\alpha - \beta) + (\alpha + \beta) = 45^\circ + 30^\circ \]

This simplifies to:

\[ 2\alpha = 75^\circ \]

Thus, we find:

\[ \alpha = \frac{75^\circ}{2} = 37.5^\circ \]

### Step 5: Solve for \( \beta \)

Now, substitute \( \alpha \) back into one of the equations to find \( \beta \):

Using \( \alpha + \beta = 30^\circ \):

\[ 37.5^\circ + \beta = 30^\circ \]

This gives:

\[ \beta = 30^\circ - 37.5^\circ = -7.5^\circ \]

Since \( \beta \) must be positive, we need to reconsider our assumption about \( \alpha + \beta \).

### Step 6: Re-evaluate the angles

Since \( \alpha + \beta = 30^\circ \) is not valid, we should consider the periodic nature of the trigonometric functions.

Instead, let's consider the next possible angle for \( \alpha + \beta \):

If we take \( \alpha + \beta = 210^\circ \) (which is \( 30^\circ + 180^\circ \)), we can solve again:

1. \( \alpha - \beta = 45^\circ \)
2. \( \alpha + \beta = 210^\circ \)

Adding these:

\[ 2\alpha = 255^\circ \implies \alpha = 127.5^\circ \]

### Conclusion

The smallest positive value of \( \alpha \) that satisfies both equations is:

\[ \alpha = 127.5^\circ \]