In triangle ABC,left(tan frac{A}{2}+tan frac{ | vedclass

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Question

(A) We know that in $	riangle ABC$,$	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,$	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$,and $\

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$\frac{2c}{a+b+c}$

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(A) We know that in $	riangle ABC$,$	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,$	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$,and $	an \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.Substituting these values into the expression:$\left(	an \frac{A}{2} + 	an \frac{B}{2}ight) 	an \frac{C}{2} = 	an \frac{A}{2} 	an \frac{C}{2} + 	an \frac{B}{2} 	an \frac{C}{2}$$= \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} + \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$= \frac{s-b}{s} + \frac{s-a}{s}$$= \frac{2s - a - b}{s}$Since $2s = a + b + c$,we have $2s - a - b = c$ and $s = \frac{a+b+c}{2}$.$= \frac{c}{\left(\frac{a+b+c}{2}ight)} = \frac{2c}{a+b+c}$.

In $\triangle ABC$,$\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right) \tan \frac{C}{2} =$

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