In triangle PQR,angle R = frac{pi}{4}. If tan | 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} + \f

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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 \left(\frac{P}{3} + \frac{Q}{3}ight) = 	an \left(\frac{\pi}{4}ight) = 1$.Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get $\frac{	an(P/3) + 	an(Q/3)}{1 - 	an(P/3)	an(Q/3)} = 1$.Since $	an(P/3)$ and $	an(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: $\frac{-b/a}{1 - c/a} = 1$.$\Rightarrow \frac{-b}{a - c} = 1$.$\Rightarrow -b = a - c$.$\Rightarrow a + b = c$.

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

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