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 where `sec alpha` and `cosec alpha` are the roots of the equation `x^2 - ax + b = 0`, we will use the properties of the roots of a quadratic equation.

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of the quadratic equation are given as:
- Root 1: \( r_1 = \sec \alpha \)
- Root 2: \( r_2 = \csc \alpha \)

2. **Sum of the Roots**:
According to Vieta's formulas, the sum of the roots of the equation \( x^2 - ax + b = 0 \) is equal to \( a \).
\[ r_1 + r_2 = \sec \alpha + \csc \alpha = a \]

3. **Product of the Roots**:
Similarly, the product of the roots is equal to \( b \).
\[ r_1 \cdot r_2 = \sec \alpha \cdot \csc \alpha = b \]

4. **Expressing in Terms of Sine and Cosine**:
Recall that:
\[ \sec \alpha = \frac{1}{\cos \alpha} \quad \text{and} \quad \csc \alpha = \frac{1}{\sin \alpha} \]
Therefore, we can rewrite the equations:
- From the sum:
\[ \sec \alpha + \csc \alpha = \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha} = a \]
- From the product:
\[ \sec \alpha \cdot \csc \alpha = \frac{1}{\cos \alpha} \cdot \frac{1}{\sin \alpha} = b \]

5. **Finding a Common Denominator**:
For the sum, we can find a common denominator:
\[ \frac{\sin \alpha + \cos \alpha}{\sin \alpha \cos \alpha} = a \]
This implies:
\[ \sin \alpha + \cos \alpha = a \sin \alpha \cos \alpha \]

6. **Substituting for Product**:
From the product equation:
\[ b = \frac{1}{\sin \alpha \cos \alpha} \implies \sin \alpha \cos \alpha = \frac{1}{b} \]

7. **Substituting Back**:
Substitute \( \sin \alpha \cos \alpha \) in the sum equation:
\[ \sin \alpha + \cos \alpha = \frac{a}{b} \]

8. **Squaring Both Sides**:
Now, square both sides:
\[ (\sin \alpha + \cos \alpha)^2 = \left(\frac{a}{b}\right)^2 \]
Expanding the left side:
\[ \sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha = \frac{a^2}{b^2} \]

9. **Using the Pythagorean Identity**:
We know that \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ 1 + 2 \sin \alpha \cos \alpha = \frac{a^2}{b^2} \]

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

11. **Rearranging**:
Multiply through by \( b^2 \):
\[ b^2 + 2b = a^2 \]
Rearranging gives:
\[ a^2 = b^2 + 2b \]

### Final Result:
Thus, we find that:
\[ a^2 = b + b^2 \]