The roots of both the equation `sin^(2) x + p sin x + q = 0` and `cos^(2) x + r cos x +s =0` and `alpha` and `beta`. Then the value of `sin ( alpha + beta ) =`

Options

A. `( 2 q s)/( q^(2) + s^(2))`
B. `( 2 p s)/( p^(2) + s^(2))`
C. `( 2pr)/(p^(2) + r^(2))`
D. `( 2qr)/( q^(2) + r^(2))`

Answer

To find the value of \(\sin(\alpha + \beta)\) given the equations \( \sin^2 x + p \sin x + q = 0 \) and \( \cos^2 x + r \cos x + s = 0 \) with roots \(\alpha\) and \(\beta\), we can follow these steps:

### Step 1: Identify the Relationships from the Equations
The roots of the first equation give us:
\[ \sin \alpha + \sin \beta = -p \quad (1) \]
and the roots of the second equation give us:
\[ \cos \alpha + \cos \beta = -r \quad (2) \]

### Step 2: Use the Product of the Roots
From the first equation, we know:
\[ \sin \alpha \sin \beta = q \quad (3) \]
From the second equation, we have:
\[ \cos \alpha \cos \beta = s \quad (4) \]

### Step 3: Multiply the Sine and Cosine Sums
We can use the identity for \(\sin(\alpha + \beta)\):
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
We can express this in terms of the sums and products we have:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]

### Step 4: Substitute from Equations (1) and (2)
Using the relationships from (1) and (2), we can express \(\sin(\alpha + \beta)\) in terms of \(p\) and \(r\):
\[ \sin(\alpha + \beta) = \sin \alpha (-r - \cos \alpha) + \cos \alpha (-p - \sin \alpha) \]

### Step 5: Use the Pythagorean Identity
We can also use the identity \(\sin^2 x + \cos^2 x = 1\) to relate \(\sin \alpha\) and \(\cos \alpha\):
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]

### Step 6: Combine and Simplify
Now, we can combine the equations to find:
\[ \sin(\alpha + \beta) = \frac{2pr}{p^2 + r^2} \]

### Conclusion
Thus, the value of \(\sin(\alpha + \beta)\) is:
\[ \sin(\alpha + \beta) = \frac{2pr}{p^2 + r^2} \]