In any triangle PQR, `angleR = pi/2`. If `tan 'P/2` and `tan 'Q/2` are the roots of the equation `ax^(2) + bx + c=0 (a ne 0)`, then show that, a+b=c.

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We are given a triangle PQR where angle R is \( \frac{\pi}{2} \) (90 degrees). The roots of the equation \( ax^2 + bx + c = 0 \) are \( \tan \frac{P}{2} \) and \( \tan \frac{Q}{2} \).

### Step 2: Use the properties of the roots
From the properties of quadratic equations, we know:
- The sum of the roots \( \tan \frac{P}{2} + \tan \frac{Q}{2} = -\frac{b}{a} \)
- The product of the roots \( \tan \frac{P}{2} \cdot \tan \frac{Q}{2} = \frac{c}{a} \)

### Step 3: Use the tangent addition formula
We can use the formula for the tangent of the sum of angles:
\[ \tan \left( \frac{P}{2} + \frac{Q}{2} \right) = \frac{\tan \frac{P}{2} + \tan \frac{Q}{2}}{1 - \tan \frac{P}{2} \tan \frac{Q}{2}} \]
Since \( P + Q + R = \pi \) and \( R = \frac{\pi}{2} \), we have \( P + Q = \frac{\pi}{2} \). Therefore, \( \frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4} \).

### Step 4: Calculate \( \tan \frac{\pi}{4} \)
We know that:
\[ \tan \frac{\pi}{4} = 1 \]
Thus,
\[ 1 = \frac{\tan \frac{P}{2} + \tan \frac{Q}{2}}{1 - \tan \frac{P}{2} \tan \frac{Q}{2}} \]

### Step 5: Substitute the values of the roots
Substituting the values from the roots:
\[ 1 = \frac{-\frac{b}{a}}{1 - \frac{c}{a}} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 1 - \frac{c}{a} = -\frac{b}{a} \]
Multiplying through by \( a \) (since \( a \neq 0 \)):
\[ a - c = -b \]
Rearranging this gives:
\[ a + b = c \]

### Conclusion
Thus, we have shown that \( a + b = c \).