If `cos(alpha+beta)=4/5; sin(alpha-beta)=5/13` and `alpha,beta` lie between `0&pi/4` then find the value of `tan2alpha`

Answer

To solve the problem step by step, we will use the given values of \( \cos(\alpha + \beta) \) and \( \sin(\alpha - \beta) \) to find \( \tan(2\alpha) \).

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

We know that:
\[ \cos(\alpha + \beta) = \frac{4}{5} \]
Using the Pythagorean theorem, we can find \( \sin(\alpha + \beta) \):
\[ \sin^2(\alpha + \beta) + \cos^2(\alpha + \beta) = 1 \]
\[ \sin^2(\alpha + \beta) + \left(\frac{4}{5}\right)^2 = 1 \]
\[ \sin^2(\alpha + \beta) + \frac{16}{25} = 1 \]
\[ \sin^2(\alpha + \beta) = 1 - \frac{16}{25} = \frac{9}{25} \]
\[ \sin(\alpha + \beta) = \frac{3}{5} \]

Now, we can find \( \tan(\alpha + \beta) \):
\[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

**Hint:** Use the Pythagorean identity to find \( \sin \) from \( \cos \).

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

We are given:
\[ \sin(\alpha - \beta) = \frac{5}{13} \]
Using the Pythagorean theorem again, we can find \( \cos(\alpha - \beta) \):
\[ \sin^2(\alpha - \beta) + \cos^2(\alpha - \beta) = 1 \]
\[ \left(\frac{5}{13}\right)^2 + \cos^2(\alpha - \beta) = 1 \]
\[ \frac{25}{169} + \cos^2(\alpha - \beta) = 1 \]
\[ \cos^2(\alpha - \beta) = 1 - \frac{25}{169} = \frac{144}{169} \]
\[ \cos(\alpha - \beta) = \frac{12}{13} \]

Now, we can find \( \tan(\alpha - \beta) \):
\[ \tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]

**Hint:** Again, apply the Pythagorean theorem to find \( \cos \) from \( \sin \).

### Step 3: Find \( \tan(2\alpha) \)

Using the identity:
\[ \tan(2\alpha) = \tan(\alpha + \beta + \alpha - \beta) = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta) \tan(\alpha - \beta)} \]
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 and denominator**

Finding a common denominator for the numerator:
\[ \frac{3}{4} = \frac{9}{12} \]
So,
\[ \tan(2\alpha) = \frac{\frac{9}{12} + \frac{5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{1 - \frac{15}{48}} \]
Calculating the denominator:
\[ 1 - \frac{15}{48} = \frac{48 - 15}{48} = \frac{33}{48} \]

**Step 5: Final calculation**

Now substituting back into the equation:
\[ \tan(2\alpha) = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{14 \cdot 48}{12 \cdot 33} = \frac{672}{396} \]
Simplifying:
\[ \tan(2\alpha) = \frac{56}{33} \]

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