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: Concept
In $\triangle PQR$, if $\angle R = 90^{\circ}$, then $P+Q = 90^{\circ}$, which means $\frac{P+Q}{2} = 45^{\circ}$.

Step 2: Analysis
Using roots of quadratic equation:
Sum of roots $S = \tan\frac{P}{2} + \tan\frac{Q}{2} = -b/a$
Product of roots $P = \tan\frac{P}{2} \tan\frac{Q}{2} = c/a$

Step 3: Calculation
$\tan(\frac{P+Q}{2}) = \tan 45^{\circ} = 1$

$\frac{\tan\frac{P}{2} + \tan\frac{Q}{2}}{1 - \tan\frac{P}{2} \tan\frac{Q}{2}} = 1 \implies \frac{-b/a}{1 - c/a} = 1$

$-b/a = (a-c)/a \implies -b = a - c \implies a + b = c$.

Step 4: Conclusion
Hence, $a+b=c$ is the correct relation. Final Answer: (A)