Let `cot alpha and cot beta` be two real roots of the equation `5 cot ^(2) x- 3 cot x-1=0,` then `cot ^(2)(alpha + beta)=`

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the problem, we need to find the value of \( \cot^2(\alpha + \beta) \) given that \( \cot \alpha \) and \( \cot \beta \) are the roots of the quadratic equation \( 5 \cot^2 x - 3 \cot x - 1 = 0 \).

### Step-by-Step Solution:

1. **Identify the coefficients of the quadratic equation:**
The equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 5 \)
- \( b = -3 \)
- \( c = -1 \)

2. **Calculate the sum of the roots:**
The sum of the roots \( \cot \alpha + \cot \beta \) can be found using the formula:
\[ \text{Sum of roots} = -\frac{b}{a} = -\frac{-3}{5} = \frac{3}{5} \]

3. **Calculate the product of the roots:**
The product of the roots \( \cot \alpha \cdot \cot \beta \) can be found using the formula:
\[ \text{Product of roots} = \frac{c}{a} = \frac{-1}{5} \]

4. **Use the cotangent addition formula:**
We know the formula for \( \cot(\alpha + \beta) \):
\[ \cot(\alpha + \beta) = \frac{\cot \alpha \cdot \cot \beta - 1}{\cot \alpha + \cot \beta} \]
Substituting the values we found:
\[ \cot(\alpha + \beta) = \frac{\left(-\frac{1}{5}\right) - 1}{\frac{3}{5}} = \frac{-\frac{1}{5} - \frac{5}{5}}{\frac{3}{5}} = \frac{-\frac{6}{5}}{\frac{3}{5}} \]

5. **Simplify the expression:**
Dividing the fractions:
\[ \cot(\alpha + \beta) = \frac{-6}{5} \cdot \frac{5}{3} = -2 \]

6. **Find \( \cot^2(\alpha + \beta) \):**
Now we need to find \( \cot^2(\alpha + \beta) \):
\[ \cot^2(\alpha + \beta) = (-2)^2 = 4 \]

### Final Answer:
Thus, the value of \( \cot^2(\alpha + \beta) \) is \( \boxed{4} \).