If `sin(alpha+beta)=(4)/(5), sin(alpha-beta)=(5)/(13)`, find the value of `tan 2alpha`:

Options

A. `(16)/(36)`
B. `(63)/(16)`
C. `(36)/(16)`
D. none of these

Answer

To find the value of \( \tan 2\alpha \) given that \( \sin(\alpha + \beta) = \frac{4}{5} \) and \( \sin(\alpha - \beta) = \frac{5}{13} \), we can follow these steps:

### Step 1: Find \( \cos(\alpha + \beta) \)
We know that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
For \( \sin(\alpha + \beta) = \frac{4}{5} \):
\[ \cos^2(\alpha + \beta) = 1 - \sin^2(\alpha + \beta) = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]
Thus,
\[ \cos(\alpha + \beta) = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Step 2: Find \( \cos(\alpha - \beta) \)
For \( \sin(\alpha - \beta) = \frac{5}{13} \):
\[ \cos^2(\alpha - \beta) = 1 - \sin^2(\alpha - \beta) = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169} \]
Thus,
\[ \cos(\alpha - \beta) = \sqrt{\frac{144}{169}} = \frac{12}{13} \]

### Step 3: Calculate \( \tan(\alpha + \beta) \) 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: Use the formula for \( \tan 2\alpha \)
We can use the formula:
\[ \tan 2\alpha = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta) \tan(\alpha - \beta)} \]
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: Simplify the numerator
To add \( \frac{4}{3} \) and \( \frac{5}{12} \), we need a common denominator:
\[ \frac{4}{3} = \frac{16}{12} \]
So,
\[ \tan 2\alpha = \frac{\frac{16}{12} + \frac{5}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{1 - \frac{5}{9}} \]

### Step 6: Simplify the denominator
Calculating \( 1 - \frac{5}{9} \):
\[ 1 - \frac{5}{9} = \frac{4}{9} \]

### Step 7: Final calculation
Now substituting back:
\[ \tan 2\alpha = \frac{\frac{21}{12}}{\frac{4}{9}} = \frac{21}{12} \cdot \frac{9}{4} = \frac{189}{48} \]
Simplifying \( \frac{189}{48} \):
\[ \tan 2\alpha = \frac{63}{16} \]

Thus, the final answer is:
\[ \tan 2\alpha = \frac{63}{16} \]