In a triangle PQR,angle R = frac{pi}{2}. If t | vedclass

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(A) In $	riangle PQR$,$\angle R = \frac{\pi}{2}$,so $P + Q = \frac{\pi}{2}$,which implies $\frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4}$.Given that $	a

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$a + b = c$

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(A) In $	riangle PQR$,$\angle R = \frac{\pi}{2}$,so $P + Q = \frac{\pi}{2}$,which implies $\frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4}$.Given that $	an(\frac{P}{2})$ and $	an(\frac{Q}{2})$ are roots of $ax^2 + bx + c = 0$,we have:Sum of roots: $	an(\frac{P}{2}) + 	an(\frac{Q}{2}) = -\frac{b}{a}$Product of roots: $	an(\frac{P}{2}) 	an(\frac{Q}{2}) = \frac{c}{a}$Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$:$	an(\frac{P}{2} + \frac{Q}{2}) = 	an(\frac{\pi}{4}) = 1$$\frac{	an(\frac{P}{2}) + 	an(\frac{Q}{2})}{1 - 	an(\frac{P}{2}) 	an(\frac{Q}{2})} = 1$Substituting the sum and product of roots:$\frac{-b/a}{1 - c/a} = 1$$\frac{-b/a}{(a-c)/a} = 1$$-b = a - c$$c = a + b$.

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

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