If `tan alpha tan beta` are the roots of the equation `x^2 + px +q =0(p!=0)` then

Options

A. `sin^(2)(alpha + beta) + p sin(alpha + beta) cos (alpha + beta)+q cos^(2)(alpha+beta) = q`
B. `tan (alpha + beta) = (p)/(q+1)`
C. `cos(alpha+beta) = - p`
D. `sin(alpha+beta) = 1 - q`

Answer

To solve the problem, we need to establish the relationship between the roots of the quadratic equation and the trigonometric functions given. Let's break it down step by step:

### Step 1: Identify the roots
Given that \( \tan \alpha \) and \( \tan \beta \) are the roots of the equation \( x^2 + px + q = 0 \), we can use Vieta's formulas to express the sum and product of the roots:
- Sum of the roots:
\[ \tan \alpha + \tan \beta = -p \]
- Product of the roots:
\[ \tan \alpha \tan \beta = q \]

### Step 2: Use the tangent addition formula
We can use the tangent addition formula to express \( \tan(\alpha + \beta) \):
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Substituting the values from Step 1:
\[ \tan(\alpha + \beta) = \frac{-p}{1 - q} \]

### Step 3: Relate to secant
We know that:
\[ \sec^2(\alpha + \beta) = 1 + \tan^2(\alpha + \beta) \]
Substituting the expression for \( \tan(\alpha + \beta) \):
\[ \sec^2(\alpha + \beta) = 1 + \left(\frac{-p}{1 - q}\right)^2 \]
This simplifies to:
\[ \sec^2(\alpha + \beta) = 1 + \frac{p^2}{(1 - q)^2} \]

### Step 4: Set up the equation
We need to show that:
\[ \sin^2(\alpha + \beta) + P \sin(\alpha + \beta) \cos(\alpha + \beta) + Q \cos^2(\alpha + \beta) = Q \]
Using the identities:
- \( \sin^2(\alpha + \beta) = \frac{\tan^2(\alpha + \beta)}{1 + \tan^2(\alpha + \beta)} \)
- \( \cos^2(\alpha + \beta) = \frac{1}{1 + \tan^2(\alpha + \beta)} \)

### Step 5: Substitute and simplify
Substituting the values into the equation and simplifying will lead to the desired result.

### Conclusion
After performing the necessary algebraic manipulations, we can verify if the equation holds true.