Let ` cos (alpha+beta) = 4/5` and `sin(alpha-beta)=5/13 ` where `0<= alpha,beta<= pi/4` then find ` tan (2alpha)`

Options

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

Explanation

To solve the problem, we need to find \( \tan(2\alpha) \) given that \( \cos(\alpha + \beta) = \frac{4}{5} \) and \( \sin(\alpha - \beta) = \frac{5}{13} \).

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

Given \( \cos(\alpha + \beta) = \frac{4}{5} \), we can find \( \sin(\alpha + \beta) \) using the Pythagorean identity:

\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]

Let \( \sin(\alpha + \beta) = y \). Then,

\[ y^2 + \left(\frac{4}{5}\right)^2 = 1 \]

Calculating \( \left(\frac{4}{5}\right)^2 \):

\[ y^2 + \frac{16}{25} = 1 \]

Subtracting \( \frac{16}{25} \) from both sides:

\[ y^2 = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \]

Taking the square root:

\[ y = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

Thus,

\[ \sin(\alpha + \beta) = \frac{3}{5} \]

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

\[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

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

Given \( \sin(\alpha - \beta) = \frac{5}{13} \), we can find \( \cos(\alpha - \beta) \) using the Pythagorean identity:

Let \( \cos(\alpha - \beta) = z \). Then,

\[ \left(\frac{5}{13}\right)^2 + z^2 = 1 \]

Calculating \( \left(\frac{5}{13}\right)^2 \):

\[ \frac{25}{169} + z^2 = 1 \]

Subtracting \( \frac{25}{169} \) from both sides:

\[ z^2 = 1 - \frac{25}{169} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169} \]

Taking the square root:

\[ z = \sqrt{\frac{144}{169}} = \frac{12}{13} \]

Thus,

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

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

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

### Step 3: Use the tangent addition formula to find \( \tan(2\alpha) \)

We know:

\[ \tan(2\alpha) = \tan((\alpha + \beta) + (\alpha - \beta)) = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta) \tan(\alpha - \beta)} \]

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

Finding a common denominator for the numerator:

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

Thus,

\[ \tan(2\alpha) = \frac{\frac{9}{12} + \frac{5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{1 - \frac{15}{48}} \]

### Step 5: Simplify the denominator

Calculating \( 1 - \frac{15}{48} \):

\[ 1 = \frac{48}{48} \implies 1 - \frac{15}{48} = \frac{48 - 15}{48} = \frac{33}{48} \]

### Step 6: Final calculation

Now substituting back into the equation:

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

Simplifying \( \frac{14 \cdot 48}{12 \cdot 33} \):

\[ = \frac{14 \cdot 4}{33} = \frac{56}{33} \]

Thus, the final answer is:

\[ \boxed{\frac{56}{33}} \]