In triangle ABC,with usual notations,m angle | vedclass

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Question

(A) In $	riangle ABC$,$\angle A + \angle B + \angle C = 180^{\circ}$.Since $\angle C = \frac{\pi}{2} = 90^{\circ}$,we have $\angle A + \angle B = 90^

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$a_1 + b_1 = c_1$

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(A) In $	riangle ABC$,$\angle A + \angle B + \angle C = 180^{\circ}$.Since $\angle C = \frac{\pi}{2} = 90^{\circ}$,we have $\angle A + \angle B = 90^{\circ}$.Therefore,$\frac{A}{2} + \frac{B}{2} = 45^{\circ} = \frac{\pi}{4}$.Given that $	an \left(\frac{A}{2}ight)$ and $	an \left(\frac{B}{2}ight)$ are the roots of $a_1 x^2 + b_1 x + c_1 = 0$,by the relation between roots and coefficients:Sum of roots: $	an \left(\frac{A}{2}ight) + 	an \left(\frac{B}{2}ight) = -\frac{b_1}{a_1}$.Product of roots: $	an \left(\frac{A}{2}ight) \cdot 	an \left(\frac{B}{2}ight) = \frac{c_1}{a_1}$.Using the formula $	an \left(\frac{A}{2} + \frac{B}{2}ight) = \frac{	an \left(\frac{A}{2}ight) + 	an \left(\frac{B}{2}ight)}{1 - 	an \left(\frac{A}{2}ight) 	an \left(\frac{B}{2}ight)}$:$	an \left(\frac{\pi}{4}ight) = \frac{-\frac{b_1}{a_1}}{1 - \frac{c_1}{a_1}}$.$1 = \frac{-b_1}{a_1 - c_1}$.$a_1 - c_1 = -b_1$,which implies $a_1 + b_1 = c_1$.

In $\triangle ABC$,with usual notations,$m \angle C = \frac{\pi}{2}$. If $\tan \left(\frac{A}{2}\right)$ and $\tan \left(\frac{B}{2}\right)$ are the roots of the equation $a_1 x^2 + b_1 x + c_1 = 0$ $(a_1 \neq 0)$,then:

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