In triangle ABC,(a+b+c)left(tan frac{A}{2}+ta | vedclass

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(A) We know that $a+b+c = 2s$,where $s$ is the semi-perimeter of the triangle.Using the formula $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

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$2c \cot \frac{C}{2}$

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(A) We know that $a+b+c = 2s$,where $s$ is the semi-perimeter of the triangle.Using the formula $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$,we can simplify the expression.Alternatively,using $	an \frac{A}{2} = \frac{\Delta}{s(s-a)}$ is not standard,but $	an \frac{A}{2} = \frac{r}{s-a}$ where $r$ is the inradius.Thus,$(a+b+c)\left(	an \frac{A}{2}+	an \frac{B}{2}ight) = 2s \left(\frac{r}{s-a} + \frac{r}{s-b}ight)$.$= 2sr \left(\frac{s-b+s-a}{(s-a)(s-b)}ight) = 2sr \left(\frac{c}{(s-a)(s-b)}ight)$.Since $r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$,we have $r^2 = \frac{(s-a)(s-b)(s-c)}{s}$.Substituting this,the expression simplifies to $2c \cot \frac{C}{2}$.

In $\triangle ABC$,$(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to

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