In a triangle PQR with usual notations,angle | vedclass

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(A) In $	riangle PQR$,$\angle R = \frac{\pi}{2}$,so $P + Q = \frac{\pi}{2}$.Thus,$\frac{P+Q}{2} = \frac{\pi}{4}$.We know that $	an \frac{P}{2}$ and

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$a + b = c$

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(A) In $	riangle PQR$,$\angle R = \frac{\pi}{2}$,so $P + Q = \frac{\pi}{2}$.Thus,$\frac{P+Q}{2} = \frac{\pi}{4}$.We know that $	an \frac{P}{2}$ and $	an \frac{Q}{2}$ are roots of $ax^2 + bx + c = 0$.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(\frac{P}{2} + \frac{Q}{2}) = \frac{	an \frac{P}{2} + 	an \frac{Q}{2}}{1 - 	an \frac{P}{2} 	an \frac{Q}{2}}$.Since $\frac{P+Q}{2} = \frac{\pi}{4}$,$	an(\frac{\pi}{4}) = 1$.So,$1 = \frac{-b/a}{1 - c/a} = \frac{-b/a}{(a-c)/a} = \frac{-b}{a-c}$.$a - c = -b$,which implies $a + b = c$.

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 $ax^2 + bx + c = 0$ $(a \neq 0)$,then:

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