If roots of the equation `x^(2) + px + q = 0` are `tan19` and `tan26` then which of the following is correct ?

Options

A. q - p = 1
B. p - q = 1
C. p + q = 2
D. p + q = 3

Answer

To solve the problem, we need to find the relationship between the coefficients \( p \) and \( q \) of the quadratic equation \( x^2 + px + q = 0 \) given that its roots are \( \tan 19^\circ \) and \( \tan 26^\circ \).

### Step-by-Step Solution:

1. **Identify the roots:**
The roots of the quadratic equation are given as \( \alpha = \tan 19^\circ \) and \( \beta = \tan 26^\circ \).

2. **Sum of the roots:**
The sum of the roots \( \alpha + \beta \) can be expressed using the formula:
\[ \alpha + \beta = -\frac{p}{1} \]
Therefore,
\[ \tan 19^\circ + \tan 26^\circ = -p \]

3. **Product of the roots:**
The product of the roots \( \alpha \beta \) is given by:
\[ \alpha \beta = \frac{q}{1} \]
Hence,
\[ \tan 19^\circ \cdot \tan 26^\circ = q \]

4. **Using the tangent addition formula:**
We can use the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
For \( A = 19^\circ \) and \( B = 26^\circ \), we have:
\[ \tan(19^\circ + 26^\circ) = \tan 45^\circ = 1 \]
Therefore,
\[ 1 = \frac{\tan 19^\circ + \tan 26^\circ}{1 - \tan 19^\circ \tan 26^\circ} \]

5. **Setting up the equation:**
From the above equation, we can cross-multiply:
\[ 1 - \tan 19^\circ \tan 26^\circ = \tan 19^\circ + \tan 26^\circ \]
Substituting the values from steps 2 and 3:
\[ 1 - q = -p \]

6. **Rearranging the equation:**
Rearranging gives us:
\[ p - q = -1 \]
or equivalently,
\[ q - p = 1 \]

7. **Conclusion:**
Therefore, the correct relationship between \( p \) and \( q \) is:
\[ q - p = 1 \]

### Final Answer:
The correct option is **A: \( q - p = 1 \)**.