Question

In a triangle PQR with usual notations, $\angle R=\frac{\pi}{2}$. If $\tan\frac{P}{2}$ and $\tan\frac{Q}{2}$ are the roots of the equation $ax^{2}+bx+c=0(a\ne0),$ then}

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

Hint:

For $P+Q = 90^{\circ}$, the sum of half-angle tangents divided by $(1 - \text{product})$ is always 1.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
In triangle $PQR$, $\angle R = 90^\circ$. We need to find a relation between $a, b, c$ given that the roots of the quadratic equation are $\tan(P/2)$ and $\tan(Q/2)$.

Step 2: Key Formula or Approach:
1. Angle sum property: $P + Q + R = \pi$.
2. Quadratic properties: Sum of roots $= -b/a$, Product of roots $= c/a$.
3. Tangent sum formula: $\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$.

Step 3: Detailed Explanation:
Since $\angle R = \frac{\pi}{2}$, we have $P + Q = \frac{\pi}{2}$, which implies $\frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4}$.
Taking tangent on both sides: \[ \tan(\frac{P}{2} + \frac{Q}{2}) = \tan \frac{\pi}{4} = 1 \] \[ \frac{\tan \frac{P}{2} + \tan \frac{Q}{2}}{1 - \tan \frac{P}{2} \tan \frac{Q}{2}} = 1 \] Let the roots be $\alpha = \tan \frac{P}{2}$ and $\beta = \tan \frac{Q}{2}$.
From the equation $ax^2 + bx + c = 0$, $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Substituting into the tangent sum formula: \[ \frac{-b/a}{1 - c/a} = 1 \] \[ \frac{-b}{a - c} = 1 \] \[ -b = a - c \] \[ a + b = c \]

Step 4: Final Answer:
The condition is $a + b = c$.