If `sin (alpha + beta) = 4/5 , cos (alpha - beta) = (12)/(13) 0 lt alpha lt (pi)/(4), 0 lt beta lt (pi)/(4),` find `tan 2 alpha .`

Answer

To solve the problem step by step, we will use the given values and trigonometric identities.

### Step 1: Find \( \cos(\alpha + \beta) \)
We know that:
\[ \sin(\alpha + \beta) = \frac{4}{5} \]
Using the identity \( \cos^2(\theta) = 1 - \sin^2(\theta) \), we can find \( \cos(\alpha + \beta) \):
\[ \cos(\alpha + \beta) = \sqrt{1 - \sin^2(\alpha + \beta)} = \sqrt{1 - \left(\frac{4}{5}\right)^2} \]
Calculating this gives:
\[ \cos(\alpha + \beta) = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 2: Find \( \sin(\alpha - \beta) \)
We are given:
\[ \cos(\alpha - \beta) = \frac{12}{13} \]
Using the identity \( \sin^2(\theta) = 1 - \cos^2(\theta) \), we can find \( \sin(\alpha - \beta) \):
\[ \sin(\alpha - \beta) = \sqrt{1 - \cos^2(\alpha - \beta)} = \sqrt{1 - \left(\frac{12}{13}\right)^2} \]
Calculating this gives:
\[ \sin(\alpha - \beta) = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \]

### Step 3: Find \( \tan(\alpha + \beta) \) and \( \tan(\alpha - \beta) \)
Using the definitions of tangent:
\[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]
\[ \tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]

### Step 4: Find \( \tan(2\alpha) \)
Using the formula for \( \tan(2\theta) \):
\[ \tan(2\alpha) = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta) \tan(\alpha - \beta)} \]
Substituting the values we found:
\[ \tan(2\alpha) = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \left(\frac{4}{3} \cdot \frac{5}{12}\right)} \]

### Step 5: Simplifying the numerator
Finding a common denominator for the numerator:
\[ \frac{4}{3} = \frac{16}{12} \]
Thus,
\[ \tan(2\alpha) = \frac{\frac{16}{12} + \frac{5}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{1 - \frac{5}{9}} \]

### Step 6: Simplifying the denominator
Calculating the denominator:
\[ 1 - \frac{5}{9} = \frac{4}{9} \]
So now we have:
\[ \tan(2\alpha) = \frac{\frac{21}{12}}{\frac{4}{9}} = \frac{21}{12} \cdot \frac{9}{4} = \frac{189}{48} = \frac{63}{16} \]

### Final Answer
Thus, the value of \( \tan(2\alpha) \) is:
\[ \boxed{\frac{63}{16}} \]