In a triangle PQR, angle R = pi / 2. If tan(P | vedclass

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(A) In a triangle $PQR$, we have $P + Q + R = \pi$. Since $\angle R = \pi / 2$, it follows that $P + Q = \pi / 2$, which implies $\frac{P}{2} + \frac{

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$c = a + b$

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(A) In a triangle $PQR$, we have $P + Q + R = \pi$. Since $\angle R = \pi / 2$, it follows that $P + Q = \pi / 2$, which implies $\frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4}$.Given that $	an(P/2)$ and $	an(Q/2)$ are roots of $ax^2 + bx + c = 0$, by Vieta's formulas:Sum of roots: $	an(P/2) + 	an(Q/2) = -b/a$Product of roots: $	an(P/2) 	an(Q/2) = c/a$Using the identity $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$, we have:$	an(\frac{P}{2} + \frac{Q}{2}) = 	an(\frac{\pi}{4}) = 1$$\frac{-b/a}{1 - c/a} = 1$$\frac{-b}{a - c} = 1$$-b = a - c$$c = a + b$

In a triangle $PQR$, $\angle R = \pi / 2$. If $\tan(P/2)$ and $\tan(Q/2)$ are the roots of the quadratic equation $ax^2 + bx + c = 0$, where $a \neq 0$, then which of the following is true?

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