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

Answer

To solve the problem, we need to find the value of \( \cos(\alpha + \beta) \) given that \( \tan\left(\frac{\alpha}{2}\right) \) and \( \tan\left(\frac{\beta}{2}\right) \) are the roots of the quadratic equation \( 8x^2 - 26x + 15 = 0 \).

### Step-by-Step Solution:

1. **Identify the coefficients from the quadratic equation**:
The given quadratic equation is \( 8x^2 - 26x + 15 = 0 \).
Here, \( a = 8 \), \( b = -26 \), and \( c = 15 \).

2. **Calculate the sum of the roots**:
The sum of the roots \( \tan\left(\frac{\alpha}{2}\right) + \tan\left(\frac{\beta}{2}\right) \) can be found using the formula:
\[ \text{Sum of roots} = -\frac{b}{a} = -\frac{-26}{8} = \frac{26}{8} = \frac{13}{4} \]

3. **Calculate the product of the roots**:
The product of the roots \( \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right) \) is given by:
\[ \text{Product of roots} = \frac{c}{a} = \frac{15}{8} \]

4. **Use the tangent addition formula**:
We know that:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\tan\left(\frac{\alpha}{2}\right) + \tan\left(\frac{\beta}{2}\right)}{1 - \tan\left(\frac{\alpha}{2}\right) \tan\left(\frac{\beta}{2}\right)} \]
Substituting the values we found:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\frac{13}{4}}{1 - \frac{15}{8}} \]

5. **Simplify the denominator**:
First, calculate \( 1 - \frac{15}{8} \):
\[ 1 - \frac{15}{8} = \frac{8}{8} - \frac{15}{8} = -\frac{7}{8} \]

6. **Substitute back into the tangent formula**:
Now substituting back:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\frac{13}{4}}{-\frac{7}{8}} = \frac{13}{4} \cdot -\frac{8}{7} = -\frac{104}{28} = -\frac{26}{7} \]

7. **Find \( \cos(\alpha + \beta) \)**:
We can use the identity:
\[ \cos(\alpha + \beta) = \frac{1 - \tan^2\left(\frac{\alpha + \beta}{2}\right)}{1 + \tan^2\left(\frac{\alpha + \beta}{2}\right)} \]
First, calculate \( \tan^2\left(\frac{\alpha + \beta}{2}\right) \):
\[ \tan^2\left(\frac{\alpha + \beta}{2}\right) = \left(-\frac{26}{7}\right)^2 = \frac{676}{49} \]

8. **Substitute into the cosine formula**:
Now substituting into the cosine formula:
\[ \cos(\alpha + \beta) = \frac{1 - \frac{676}{49}}{1 + \frac{676}{49}} = \frac{\frac{49 - 676}{49}}{\frac{49 + 676}{49}} = \frac{49 - 676}{49 + 676} = \frac{-627}{725} \]

Thus, the value of \( \cos(\alpha + \beta) \) is:
\[ \cos(\alpha + \beta) = -\frac{627}{725} \]