If `alpha and beta` are the roots of the equation `x^2 + 5x-49=0`, then find the value of `cot(cot^-1 alpha + cot^-1 beta)`.

Answer

To solve the problem, we need to find the value of \( \cot(\cot^{-1} \alpha + \cot^{-1} \beta) \), where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 + 5x - 49 = 0 \).

### Step 1: Find the roots \( \alpha \) and \( \beta \)

The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \( x^2 + 5x - 49 = 0 \):
- \( a = 1 \)
- \( b = 5 \)
- \( c = -49 \)

Calculating the discriminant:
\[ b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (-49) = 25 + 196 = 221 \]

Now, substituting into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{221}}{2} \]
Thus, the roots are:
\[ \alpha = \frac{-5 + \sqrt{221}}{2}, \quad \beta = \frac{-5 - \sqrt{221}}{2} \]

### Step 2: Use the cotangent addition formula

We need to find \( \cot(\cot^{-1} \alpha + \cot^{-1} \beta) \). We can use the formula:
\[ \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \]
Here, \( A = \cot^{-1} \alpha \) and \( B = \cot^{-1} \beta \). Thus:
\[ \cot A = \alpha, \quad \cot B = \beta \]

Substituting into the formula:
\[ \cot(\cot^{-1} \alpha + \cot^{-1} \beta) = \frac{\alpha \beta - 1}{\alpha + \beta} \]

### Step 3: Calculate \( \alpha + \beta \) and \( \alpha \beta \)

From Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -5 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = -49 \)

### Step 4: Substitute the values into the cotangent formula

Now substituting \( \alpha + \beta \) and \( \alpha \beta \) into the cotangent formula:
\[ \cot(\cot^{-1} \alpha + \cot^{-1} \beta) = \frac{-49 - 1}{-5} = \frac{-50}{-5} = 10 \]

### Final Answer

Thus, the value of \( \cot(\cot^{-1} \alpha + \cot^{-1} \beta) \) is:
\[ \boxed{10} \]