If `cos(alpha+beta)=(4)/(5)` and `sin(alpha-beta)=(5)/(13) and alpha, beta` lie between `0^(@) and 45^(@)`, find the value of `tan 2alpha`:

Options

A. `(56)/(63)`
B. `(56)/(33)`
C. `(25)/(33)`
D. `(65)/(63)`

Answer

To find the value of \( \tan 2\alpha \), we will use the given information:

1. \( \cos(\alpha + \beta) = \frac{4}{5} \)
2. \( \sin(\alpha - \beta) = \frac{5}{13} \)

We will follow these steps:

### Step 1: Find \( \tan(\alpha + \beta) \)

Using the identity \( \cos(\theta) = \frac{\text{base}}{\text{hypotenuse}} \), we can create a right triangle for \( \alpha + \beta \).

- Hypotenuse = 5
- Base = 4

To find the perpendicular side, we use the Pythagorean theorem:

\[ \text{Perpendicular} = \sqrt{(\text{Hypotenuse})^2 - (\text{Base})^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \]

Now, we can find \( \tan(\alpha + \beta) \):

\[ \tan(\alpha + \beta) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4} \]

### Step 2: Find \( \tan(\alpha - \beta) \)

Using the identity \( \sin(\theta) = \frac{\text{perpendicular}}{\text{hypotenuse}} \), we can create a right triangle for \( \alpha - \beta \).

- Hypotenuse = 13
- Perpendicular = 5

To find the base side, we again use the Pythagorean theorem:

\[ \text{Base} = \sqrt{(\text{Hypotenuse})^2 - (\text{Perpendicular})^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \]

Now, we can find \( \tan(\alpha - \beta) \):

\[ \tan(\alpha - \beta) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{5}{12} \]

### Step 3: 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:

\[ \tan(2\alpha) = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \left(\frac{3}{4} \cdot \frac{5}{12}\right)} \]

### Step 4: Simplify the numerator

To add \( \frac{3}{4} \) and \( \frac{5}{12} \), we need a common denominator. The least common multiple of 4 and 12 is 12.

\[ \frac{3}{4} = \frac{9}{12} \]

Now we can add:

\[ \frac{9}{12} + \frac{5}{12} = \frac{14}{12} \]

### Step 5: Simplify the denominator

Now calculate the denominator:

\[ 1 - \left(\frac{3}{4} \cdot \frac{5}{12}\right) = 1 - \frac{15}{48} \]

Convert 1 to have a denominator of 48:

\[ 1 = \frac{48}{48} \]

Now, subtract:

\[ \frac{48}{48} - \frac{15}{48} = \frac{33}{48} \]

### Step 6: Combine the results

Now we have:

\[ \tan(2\alpha) = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{14}{12} \cdot \frac{48}{33} \]

### Step 7: Simplify the fraction

Calculating this gives:

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

Now simplify \( \frac{672}{396} \):

\[ \frac{672 \div 12}{396 \div 12} = \frac{56}{33} \]

Thus, the final answer is:

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