If `secalpha` and `cosecalpha` are the roots of the equation `x^2-px+q=0`, then (i)`p^2=q(q-2)` (ii)`p^2=q(q+2)` (iii)`p^2+q^2=2q` (iv) none of these

Options

A. `p^2=q(q-2)`
B. `p^2=q(q+2)`
C. `p^2+q^2=2q`
D. None of these

Explanation

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

### Step-by-Step Solution:

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

2. **Sum of the Roots**:
According to Vieta's formulas, the sum of the roots 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. **Expressing Roots in Terms of Sine and Cosine**:
We know:
\[ \sec \alpha = \frac{1}{\cos \alpha} \quad \text{and} \quad \csc \alpha = \frac{1}{\sin \alpha} \]
Therefore, we can express the sum and product of the roots as:
\[ \sec \alpha + \csc \alpha = \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha} = p \]
\[ \sec \alpha \cdot \csc \alpha = \frac{1}{\cos \alpha \sin \alpha} = q \]

5. **Finding the Sum**:
To combine the terms in the sum:
\[ \sec \alpha + \csc \alpha = \frac{\sin \alpha + \cos \alpha}{\sin \alpha \cos \alpha} = p \]

6. **Finding the Product**:
The product can be rewritten as:
\[ \sec \alpha \cdot \csc \alpha = \frac{1}{\sin \alpha \cos \alpha} = q \]

7. **Relating \( p \) and \( q \)**:
From the product, we have:
\[ \sin \alpha \cos \alpha = \frac{1}{q} \]
Substitute this into the expression for \( p \):
\[ p = \frac{\sin \alpha + \cos \alpha}{\frac{1}{q}} = q(\sin \alpha + \cos \alpha) \]

8. **Squaring the Sum**:
Now, square both sides:
\[ p^2 = q^2(\sin \alpha + \cos \alpha)^2 \]
Expanding the right-hand side:
\[ p^2 = q^2(\sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos \alpha) \]
Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we have:
\[ p^2 = q^2(1 + 2\sin \alpha \cos \alpha) \]
Substitute \( \sin \alpha \cos \alpha = \frac{1}{q} \):
\[ p^2 = q^2\left(1 + \frac{2}{q}\right) = q^2 + 2q \]

9. **Final Relation**:
Rearranging gives:
\[ p^2 = q(q + 2) \]

### Conclusion:
Thus, the correct relation is:
\[ p^2 = q(q + 2) \]