Let `cos(alpha+beta)=(4)/(5)` and let `sin(alpha-beta)=(5)/(13)`, where `0 le alpha`, `beta=(pi)/(4)`. Then`tan2alpha=`

Options

A. `(56)/(33)`
B. `(19)/(12)`
C. `(20)/(7)`
D. `(25)/(16)`

Answer

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

### Step 1: Identify the given values
We have:
- \( \cos(\alpha + \beta) = \frac{4}{5} \)
- \( \sin(\alpha - \beta) = \frac{5}{13} \)
- \( \beta = \frac{\pi}{4} \)

### Step 2: Find \( \sin(\alpha + \beta) \) and \( \cos(\alpha - \beta) \)
Using the Pythagorean identity:
\[ \sin^2(x) + \cos^2(x) = 1 \]

For \( \cos(\alpha + \beta) = \frac{4}{5} \):
\[ \sin(\alpha + \beta) = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

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

### Step 3: Use the tangent addition formula
We know:
\[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

\[ \tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]

### Step 4: Use the tangent double angle formula
We can express \( \tan(2\alpha) \) using the formula:
\[ \tan(2\alpha) = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta) \tan(\alpha - \beta)} \]

Substituting the values we found:
\[ \tan(2\alpha) = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \left(\frac{3}{4} \cdot \frac{5}{12}\right)} \]

### Step 5: Simplify the expression
First, find a common denominator for the numerator:
\[ \frac{3}{4} = \frac{9}{12} \quad \text{(common denominator is 12)} \]
So,
\[ \tan(2\alpha) = \frac{\frac{9}{12} + \frac{5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{1 - \frac{15}{48}} \]

Now simplify the denominator:
\[ 1 - \frac{15}{48} = \frac{48 - 15}{48} = \frac{33}{48} \]

Thus,
\[ \tan(2\alpha) = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{14 \cdot 48}{12 \cdot 33} = \frac{672}{396} = \frac{56}{33} \]

### Final Answer
\[ \tan(2\alpha) = \frac{56}{33} \]