Q. If `tan alpha and tan beta` are the roots of equation `x^2-12x-3=0`, then the value of `sin^2(alpha+beta)+2sin(alpha+beta)cos(alpha+beta)+5cos^2(alpha+beta)` is

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the roots of the polynomial
The given polynomial is:
\[ x^2 - 12x - 3 = 0 \]

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \( \tan \alpha + \tan \beta = -\frac{b}{a} = \frac{12}{1} = 12 \)
- The product of the roots \( \tan \alpha \tan \beta = \frac{c}{a} = \frac{-3}{1} = -3 \)

### Step 3: Use the tangent addition formula
We know that:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Substituting the values we found:
\[ \tan(\alpha + \beta) = \frac{12}{1 - (-3)} = \frac{12}{4} = 3 \]

### Step 4: Create a right triangle
To find \( \sin(\alpha + \beta) \) and \( \cos(\alpha + \beta) \), we can create a right triangle where:
- Opposite side (perpendicular) = 3 (since \( \tan(\alpha + \beta) = 3 \))
- Adjacent side (base) = 1

### Step 5: Calculate the hypotenuse
Using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{(3^2 + 1^2)} = \sqrt{9 + 1} = \sqrt{10} \]

### Step 6: Calculate \( \sin(\alpha + \beta) \) and \( \cos(\alpha + \beta) \)
Now we can find:
- \( \sin(\alpha + \beta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{\sqrt{10}} \)
- \( \cos(\alpha + \beta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{10}} \)

### Step 7: Substitute into the expression
We need to evaluate:
\[ \sin^2(\alpha + \beta) + 2\sin(\alpha + \beta)\cos(\alpha + \beta) + 5\cos^2(\alpha + \beta) \]

Calculating each term:
1. \( \sin^2(\alpha + \beta) = \left(\frac{3}{\sqrt{10}}\right)^2 = \frac{9}{10} \)
2. \( 2\sin(\alpha + \beta)\cos(\alpha + \beta) = 2 \cdot \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5} \)
3. \( 5\cos^2(\alpha + \beta) = 5 \cdot \left(\frac{1}{\sqrt{10}}\right)^2 = 5 \cdot \frac{1}{10} = \frac{5}{10} = \frac{1}{2} \)

### Step 8: Combine the results
Now we add all the terms together:
\[ \frac{9}{10} + \frac{6}{10} + \frac{5}{10} = \frac{20}{10} = 2 \]

### Final Answer
Thus, the value of \( \sin^2(\alpha + \beta) + 2\sin(\alpha + \beta)\cos(\alpha + \beta) + 5\cos^2(\alpha + \beta) \) is:
\[ \boxed{2} \]