If `sec alpha and cosec alpha` are the roots of the equation `x^(2)-ax+b=0,` then

Options

A. `a^(2)=b(b-2)`
B. `a^(2)=b(b+2)`
C. `a^(2)+b^(2)=2b`
D. none of these

Answer

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

### Step-by-Step Solution:

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

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

3. **Use the Product of Roots**: Similarly, the product of the roots is given by:
\[ \sec \alpha \cdot \csc \alpha = b \]

4. **Express in Terms of Sine and Cosine**:
- Recall the definitions:
\[ \sec \alpha = \frac{1}{\cos \alpha}, \quad \csc \alpha = \frac{1}{\sin \alpha} \]
- Substitute these into the equations:
\[ \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha} = a \]
\[ \frac{1}{\cos \alpha} \cdot \frac{1}{\sin \alpha} = b \]

5. **Combine the Sum of Roots**:
- Rewrite the sum:
\[ \frac{\sin \alpha + \cos \alpha}{\sin \alpha \cos \alpha} = a \]
- This implies:
\[ \sin \alpha + \cos \alpha = a \cdot \sin \alpha \cos \alpha \]

6. **Substitute Product of Roots**:
- From the product, we have:
\[ \sin \alpha \cos \alpha = \frac{1}{b} \]
- Substitute this into the previous equation:
\[ \sin \alpha + \cos \alpha = a \cdot \frac{1}{b} \]

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

8. **Substituting for \( \sin \alpha \cos \alpha \)**:
- Substitute \( \sin \alpha \cos \alpha = \frac{1}{2b} \):
\[ 1 + 2 \cdot \frac{1}{2b} = \frac{a^2}{b^2} \]
- Simplifying gives:
\[ 1 + \frac{1}{b} = \frac{a^2}{b^2} \]

9. **Clear the Denominator**:
- Multiply through by \( b^2 \):
\[ b^2 + b = a^2 \]

10. **Final Relation**:
- Rearranging gives us the final relation:
\[ a^2 = b^2 + b \]