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

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(B) We know that $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{\Delta}{s(s-a)}$ and $	an \frac{B}{2} = \frac{\Delta}{s(s-b)}$.Adding t

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

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(B) We know that $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{\Delta}{s(s-a)}$ and $	an \frac{B}{2} = \frac{\Delta}{s(s-b)}$.Adding these,we get $	an \frac{A}{2} + 	an \frac{B}{2} = \frac{\Delta}{s} \left( \frac{1}{s-a} + \frac{1}{s-b} ight)$.$= \frac{\Delta}{s} \left( \frac{s-b+s-a}{(s-a)(s-b)} ight) = \frac{\Delta}{s} \left( \frac{2s-a-b}{(s-a)(s-b)} ight)$.Since $2s = a+b+c$,we have $2s-a-b = c$.So,$	an \frac{A}{2} + 	an \frac{B}{2} = \frac{c \Delta}{s(s-a)(s-b)}$.Using $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,we have $\frac{\Delta}{(s-a)(s-b)} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} = \cot \frac{C}{2}$.Thus,$	an \frac{A}{2} + 	an \frac{B}{2} = \frac{c}{s} \cot \frac{C}{2} = \frac{2c}{2s} \cot \frac{C}{2} = \frac{2c \cot \frac{C}{2}}{a+b+c}$.

In $\triangle ABC$,$\tan \frac{A}{2} + \tan \frac{B}{2} =$

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