If `secalpha, "cosec"alpha` are roots of equation `x^(2) + px + q=0`, then

Options

A. a)`p^(2)=p + 2q`
B. b)`q^(2)= p + 2q`
C. c)`p^(2) = q(q+2)`
D. d)`q^(2)=p(p+2)`

Answer

To solve the problem, we need to find the relationship between the roots of the quadratic equation \( x^2 + px + q = 0 \) given that the roots are \( \sec \alpha \) and \( \csc \alpha \).

### Step-by-Step Solution:

1. **Identify the Roots**: The roots of the equation are given as \( \sec \alpha \) and \( \csc \alpha \).

2. **Sum of the Roots**: According to Vieta's formulas, the sum of the roots of the quadratic equation \( x^2 + px + q = 0 \) is given by:
\[ \sec \alpha + \csc \alpha = -p \]

3. **Product of the Roots**: The product of the roots is given by:
\[ \sec \alpha \cdot \csc \alpha = q \]

4. **Express in Terms of Sine and Cosine**: We can express \( \sec \alpha \) and \( \csc \alpha \) in terms of sine and cosine:
\[ \sec \alpha = \frac{1}{\cos \alpha}, \quad \csc \alpha = \frac{1}{\sin \alpha} \]

5. **Sum of the Roots in Terms of Sine and Cosine**:
\[ \sec \alpha + \csc \alpha = \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha} = \frac{\sin \alpha + \cos \alpha}{\sin \alpha \cos \alpha} \]
Setting this equal to \(-p\):
\[ \frac{\sin \alpha + \cos \alpha}{\sin \alpha \cos \alpha} = -p \]

6. **Product of the Roots in Terms of Sine and Cosine**:
\[ \sec \alpha \cdot \csc \alpha = \frac{1}{\cos \alpha} \cdot \frac{1}{\sin \alpha} = \frac{1}{\sin \alpha \cos \alpha} = q \]

7. **Rearranging the Equations**: From the product equation, we have:
\[ \sin \alpha \cos \alpha = \frac{1}{q} \]
Substitute this into the sum equation:
\[ \sin \alpha + \cos \alpha = -p \cdot \sin \alpha \cos \alpha = -p \cdot \frac{1}{q} \]

8. **Square Both Sides**: Now, square both sides:
\[ (\sin \alpha + \cos \alpha)^2 = \left(-\frac{p}{q}\right)^2 \]
Expanding the left side:
\[ \sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha = \frac{p^2}{q^2} \]
Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ 1 + 2 \sin \alpha \cos \alpha = \frac{p^2}{q^2} \]

9. **Substituting \( \sin \alpha \cos \alpha \)**:
\[ 1 + 2 \cdot \frac{1}{2q^2} = \frac{p^2}{q^2} \]
This simplifies to:
\[ 1 + \frac{1}{q^2} = \frac{p^2}{q^2} \]

10. **Final Rearrangement**: Multiply through by \( q^2 \):
\[ q^2 + 1 = p^2 \]

Thus, we have derived the relationship:
\[ p^2 = q^2 + 1 \]