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

Options

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

Explanation

To solve the equation \( 5 \cot^2 x - 3 \cot x - 1 = 0 \) and find \( \cot^2(\alpha + \beta) \) where \( \alpha \) and \( \beta \) are the roots, we can follow these steps:

### Step 1: Identify the coefficients
The given equation can be compared to the standard form of a quadratic equation \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = 5 \)
- \( b = -3 \)
- \( c = -1 \)

### Step 2: Calculate the sum and product of the roots
Using the relationships for the sum and product of the roots of a quadratic equation:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-3}{5} = \frac{3}{5} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{-1}{5} \)

### Step 3: Express cotangent in terms of the roots
Let:
- \( \cot \alpha + \cot \beta = S = \frac{3}{5} \)
- \( \cot \alpha \cdot \cot \beta = P = -\frac{1}{5} \)

### Step 4: Use the cotangent addition formula
The cotangent addition formula states:
\[ \cot(\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} \]
Substituting the values of \( S \) and \( P \):
\[ \cot(\alpha + \beta) = \frac{P - 1}{S} = \frac{-\frac{1}{5} - 1}{\frac{3}{5}} \]

### Step 5: Simplify the expression
First, simplify \( P - 1 \):
\[ P - 1 = -\frac{1}{5} - \frac{5}{5} = -\frac{6}{5} \]
Now substitute this back into the cotangent formula:
\[ \cot(\alpha + \beta) = \frac{-\frac{6}{5}}{\frac{3}{5}} = -\frac{6}{5} \cdot \frac{5}{3} = -2 \]

### Step 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 \( 4 \).

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