If `sin alpha and cos alpha` are roots of the equation `px^2 + qx+r=0` then :

Options

A. `p^(2) - q^(2) + 2pr = 0`
B. `(p+r)^(2) = q^(2) - r^(2)`
C. `p^(2) + q^(2) - 2 pr = 0`
D. `(p-r)^(2) = q^(2) + r^(2)`

Answer

To solve the problem where `sin alpha` and `cos alpha` are the roots of the quadratic equation `px^2 + qx + r = 0`, we can follow these steps:

### Step 1: Use the properties of roots of a quadratic equation
For a quadratic equation of the form `ax^2 + bx + c = 0`, if `x1` and `x2` are the roots, then:
- The sum of the roots \( x1 + x2 = -\frac{b}{a} \)
- The product of the roots \( x1 \cdot x2 = \frac{c}{a} \)

Here, we have:
- Roots: \( x1 = \sin \alpha \) and \( x2 = \cos \alpha \)
- Coefficients: \( a = p \), \( b = q \), and \( c = r \)

### Step 2: Calculate the sum of the roots
Using the sum of the roots:
\[ \sin \alpha + \cos \alpha = -\frac{q}{p} \]

### Step 3: Calculate the product of the roots
Using the product of the roots:
\[ \sin \alpha \cdot \cos \alpha = \frac{r}{p} \]

### Step 4: Use the Pythagorean identity
We know from trigonometry that:
\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]

### Step 5: Express the sum of squares
To relate the sum of the roots to the product, we can manipulate the identity:
\[ (\sin \alpha + \cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos \alpha \]
Substituting the known values:
\[ \left(-\frac{q}{p}\right)^2 = 1 + 2\left(\frac{r}{p}\right) \]

### Step 6: Simplify the equation
Expanding and rearranging gives:
\[ \frac{q^2}{p^2} = 1 + \frac{2r}{p} \]
Multiplying through by \( p^2 \) to eliminate the fraction:
\[ q^2 = p^2 + 2rp \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ p^2 - q^2 + 2rp = 0 \]

### Final Result
Thus, we have derived the relationship:
\[ p^2 - q^2 + 2rp = 0 \]