If `"tan"(alpha)/(2)` and `"tan"(beta)/(2)` are the roots of the equation `8x^(2)-26x+15=0`, then find the value of `"tan"((alpha+beta)/(2))`

Answer

To find the value of \(\tan\left(\frac{\alpha + \beta}{2}\right)\) given that \(\frac{\tan(\alpha)}{2}\) and \(\frac{\tan(\beta)}{2}\) are the roots of the equation \(8x^2 - 26x + 15 = 0\), we can follow these steps:

### Step 1: Identify the roots
Let:
\[ x_1 = \frac{\tan(\alpha)}{2}, \quad x_2 = \frac{\tan(\beta)}{2} \]
The roots of the quadratic equation \(8x^2 - 26x + 15 = 0\) are \(x_1\) and \(x_2\).

### Step 2: Use the sum and product of roots
From Vieta's formulas, we know:
- The sum of the roots \(x_1 + x_2 = -\frac{b}{a} = \frac{26}{8} = \frac{13}{4}\)
- The product of the roots \(x_1 \cdot x_2 = \frac{c}{a} = \frac{15}{8}\)

### Step 3: Substitute the roots
Substituting the values of \(x_1\) and \(x_2\):
\[ \frac{\tan(\alpha)}{2} + \frac{\tan(\beta)}{2} = \frac{13}{4} \]
Multiplying through by 2 gives:
\[ \tan(\alpha) + \tan(\beta) = \frac{13}{2} \]

### Step 4: Use the product of roots
From the product of the roots:
\[ \frac{\tan(\alpha)}{2} \cdot \frac{\tan(\beta)}{2} = \frac{15}{8} \]
Multiplying through by 4 gives:
\[ \tan(\alpha) \tan(\beta) = \frac{15}{2} \]

### Step 5: Use the tangent addition formula
We need to find \(\tan\left(\frac{\alpha + \beta}{2}\right)\). We can use the formula:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)} \]
Substituting the values we found:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\frac{13}{2}}{1 - \frac{15}{2}} \]

### Step 6: Simplify the expression
Calculating the denominator:
\[ 1 - \frac{15}{2} = \frac{2}{2} - \frac{15}{2} = \frac{-13}{2} \]
Now substituting back:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\frac{13}{2}}{\frac{-13}{2}} = -1 \]

### Final Answer
Thus, the value of \(\tan\left(\frac{\alpha + \beta}{2}\right)\) is:
\[ \boxed{-1} \]