Question

With usual notations in \(\triangle \mathrm{ABC}\), if \(\angle \mathrm{B}=\frac{\pi}{2}\), and \(\tan \frac{\mathrm{A}}{2}, \tan \frac{\mathrm{C}}{2}\) are roots of equation \(\mathrm{p} x^2+\mathrm{qx}+\mathrm{r}=0\), \(\mathrm{p} \neq 0\), then

• A. \(\mathrm{p}+\mathrm{q}=\mathrm{r}\)
• B. \(\mathrm{r}+\mathrm{p}=\mathrm{q}\)
• C. \(\mathrm{r}=\mathrm{p}\)
• D. \(\mathrm{p}=\mathrm{q}\)

Hint:

Since A + C = 90 degrees, use the tangent addition formula with the sum and product of roots.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
In the triangle \(B=\dfrac{\pi}{2}\), so \(A+C=\dfrac{\pi}{2}\) and \(\dfrac{A}{2}+\dfrac{C}{2}=\dfrac{\pi}{4}\).

Step 2: Use the tangent addition formula
\(\tan\left(\dfrac{A}{2}+\dfrac{C}{2}\right)=\dfrac{\tan\frac{A}{2}+\tan\frac{C}{2}}{1-\tan\frac{A}{2}\tan\frac{C}{2}}=\tan\dfrac{\pi}{4}=1\).

Step 3: Sum and product of roots
Sum \(=-\dfrac{q}{p}\), product \(=\dfrac{r}{p}\).
So \(\dfrac{-q/p}{1-r/p}=1\), giving \(-q=p-r\).

Step 4: Result
Hence \(p+q=r\).

Final Answer:
The required relation is p + q = r. \[ \boxed{\text{(A) }\mathrm{p}+\mathrm{q}=\mathrm{r}} \]