If triangle ABC is right-angled at A and tan | vedclass

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(B) In a triangle $ABC$ right-angled at $A$,we have $B+C = 90^\circ$,so $\frac{B+C}{2} = 45^\circ$.Since $	an \frac{B}{2}$ and $	an \frac{C}{2}$ are

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$a+b=c$

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(B) In a triangle $ABC$ right-angled at $A$,we have $B+C = 90^\circ$,so $\frac{B+C}{2} = 45^\circ$.Since $	an \frac{B}{2}$ and $	an \frac{C}{2}$ are roots of $ax^2+bx+c=0$,the sum of roots is $	an \frac{B}{2} + 	an \frac{C}{2} = -\frac{b}{a}$ and the product of roots is $	an \frac{B}{2} 	an \frac{C}{2} = \frac{c}{a}$.Using the identity $	an(\frac{B}{2} + \frac{C}{2}) = \frac{	an \frac{B}{2} + 	an \frac{C}{2}}{1 - 	an \frac{B}{2} 	an \frac{C}{2}}$,we substitute $	an 45^\circ = 1$:$1 = \frac{-b/a}{1 - c/a} = \frac{-b/a}{(a-c)/a} = \frac{-b}{a-c}$.This simplifies to $a-c = -b$,which rearranges to $a+b=c$.

If triangle $ABC$ is right-angled at $A$ and $\tan \frac{B}{2}, \tan \frac{C}{2}$ are roots of the equation $ax^2 + bx + c = 0$ $(a \neq 0)$,then:

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