In triangle PQR,if angle R = frac{pi}{4} and | vedclass

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(A) Given,$R = \frac{\pi}{4}$.Since $P+Q+R = \pi$,we have $P+Q = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Dividing by $3$,we get $\frac{P}{3} + \frac{Q}{

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$a+b=c$

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(A) Given,$R = \frac{\pi}{4}$.Since $P+Q+R = \pi$,we have $P+Q = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Dividing by $3$,we get $\frac{P}{3} + \frac{Q}{3} = \frac{\pi}{4}$.Taking tangent on both sides,$	an(\frac{P}{3} + \frac{Q}{3}) = 	an(\frac{\pi}{4}) = 1$.Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get $\frac{	an(\frac{P}{3}) + 	an(\frac{Q}{3})}{1 - 	an(\frac{P}{3})	an(\frac{Q}{3})} = 1$.Since $	an(\frac{P}{3})$ and $	an(\frac{Q}{3})$ are roots of $ax^2 + bx + c = 0$,the sum of roots is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$.Substituting these into the equation: $\frac{-b/a}{1 - c/a} = 1$.This simplifies to $\frac{-b}{a-c} = 1$,which implies $-b = a - c$,or $a+b = c$.

In $\triangle PQR$,if $\angle R = \frac{\pi}{4}$ and $\tan(\frac{P}{3})$,$\tan(\frac{Q}{3})$ are the roots of the equation $ax^2 + bx + c = 0$,then:

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