With usual notations in triangle ABC,if angle | vedclass

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Question

(A) In $	riangle ABC$,since $\angle B = \frac{\pi}{2}$,we have $A + C = \frac{\pi}{2}$,which implies $\frac{A}{2} + \frac{C}{2} = \frac{\pi}{4}$.Taki

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$p + q = r$

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(A) In $	riangle ABC$,since $\angle B = \frac{\pi}{2}$,we have $A + C = \frac{\pi}{2}$,which implies $\frac{A}{2} + \frac{C}{2} = \frac{\pi}{4}$.Taking tangent on both sides,$	an(\frac{A}{2} + \frac{C}{2}) = 	an(\frac{\pi}{4}) = 1$.Using the formula $	an(x+y) = \frac{	an x + 	an y}{1 - 	an x 	an y}$,we get $\frac{	an \frac{A}{2} + 	an \frac{C}{2}}{1 - 	an \frac{A}{2} 	an \frac{C}{2}} = 1$.Let $\alpha = 	an \frac{A}{2}$ and $\beta = 	an \frac{C}{2}$ be the roots of $px^2 + qx + r = 0$.From the properties of roots,$\alpha + \beta = -\frac{q}{p}$ and $\alpha \beta = \frac{r}{p}$.Substituting these into the equation $\alpha + \beta = 1 - \alpha \beta$,we get $-\frac{q}{p} = 1 - \frac{r}{p}$.Multiplying by $p$,we get $-q = p - r$,which simplifies to $r = p + q$ or $p + q = r$.

With usual notations in $\triangle ABC$,if $\angle B = \frac{\pi}{2}$,and $\tan \frac{A}{2}, \tan \frac{C}{2}$ are roots of the equation $px^2 + qx + r = 0$,$p \neq 0$,then:

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