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

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(A) In $	riangle PQR$,$\angle P + \angle Q + \angle R = 180^{\circ}$.Since $\angle R = \frac{\pi}{2} = 90^{\circ}$,we have $\angle P + \angle Q = 90^

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$a+b=c$

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(A) In $	riangle PQR$,$\angle P + \angle Q + \angle R = 180^{\circ}$.Since $\angle R = \frac{\pi}{2} = 90^{\circ}$,we have $\angle P + \angle Q = 90^{\circ}$.Dividing by $2$,we get $\frac{P}{2} + \frac{Q}{2} = 45^{\circ} = \frac{\pi}{4}$.Given that $	an \left(\frac{P}{2}ight)$ and $	an \left(\frac{Q}{2}ight)$ are roots of $ax^2 + bx + c = 0$,by Vieta's formulas:Sum of roots: $	an \left(\frac{P}{2}ight) + 	an \left(\frac{Q}{2}ight) = -\frac{b}{a}$.Product of roots: $	an \left(\frac{P}{2}ight) 	an \left(\frac{Q}{2}ight) = \frac{c}{a}$.Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$:$	an \left(\frac{P}{2} + \frac{Q}{2}ight) = \frac{	an \frac{P}{2} + 	an \frac{Q}{2}}{1 - 	an \frac{P}{2} 	an \frac{Q}{2}}$.Substituting the values: $	an \left(\frac{\pi}{4}ight) = \frac{-b/a}{1 - c/a}$.$1 = \frac{-b/a}{(a-c)/a} = \frac{-b}{a-c}$.$a - c = -b$,which implies $a + b = c$.

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

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