In a triangle `PQR, angleR = pi //2`. If tan (P/2) and tan (Q/2) are the roots of the equations `ax^(2) + bx + c = 0` where `a ne 0` , then

Options

A. a + b = c
B. b + c = a
C. a + c = b
D. b = c

Answer

To solve the problem, we need to find a condition involving the coefficients \( a \), \( b \), and \( c \) of the quadratic equation \( ax^2 + bx + c = 0 \) given that \( \tan\left(\frac{P}{2}\right) \) and \( \tan\left(\frac{Q}{2}\right) \) are its roots.

### Step-by-Step Solution:

1. **Understanding the Triangle**:
Given that triangle \( PQR \) has \( \angle R = \frac{\pi}{2} \), we know that \( P + Q = \frac{\pi}{2} \).

2. **Using the Half-Angle Tangent Identity**:
The sum of the angles \( P \) and \( Q \) leads us to use the identity for tangent:
\[ \tan\left(\frac{P + Q}{2}\right) = \tan\left(\frac{\frac{\pi}{2}}{2}\right) = \tan\left(\frac{\pi}{4}\right) = 1 \]

3. **Sum and Product of Roots**:
If \( \tan\left(\frac{P}{2}\right) \) and \( \tan\left(\frac{Q}{2}\right) \) are the roots of the quadratic equation \( ax^2 + bx + c = 0 \), then:
- The sum of the roots \( \tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right) = -\frac{b}{a} \)
- The product of the roots \( \tan\left(\frac{P}{2}\right) \tan\left(\frac{Q}{2}\right) = \frac{c}{a} \)

4. **Using the Tangent Addition Formula**:
We can express the sum of the tangents using the tangent addition formula:
\[ \tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right) = \frac{\tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right)}{1 - \tan\left(\frac{P}{2}\right) \tan\left(\frac{Q}{2}\right)} \]
Since \( P + Q = \frac{\pi}{2} \), we have:
\[ \tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right) = 1 - \tan\left(\frac{P}{2}\right) \tan\left(\frac{Q}{2}\right) \]

5. **Setting Up the Equations**:
From the above, we can set up the equations:
- \( -\frac{b}{a} = 1 - \frac{c}{a} \)
- Rearranging gives us \( b + c = a \).

6. **Final Condition**:
Therefore, the condition involving \( a \), \( b \), and \( c \) is:
\[ a + b = c \]