In triangle ABC with usual notations, if 1 + | vedclass

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(C) We know that $	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)}}$.Multiplying these,

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$a + b$

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(C) We know that $	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)}}$.Multiplying these, we get $	an(\frac{A}{2}) 	an(\frac{B}{2}) = \sqrt{\frac{(s-b)(s-c)(s-a)(s-c)}{s(s-a)s(s-b)}} = \sqrt{\frac{(s-c)^2}{s^2}} = \frac{s-c}{s}$.Substituting this into the given expression: $1 + \frac{s-c}{s} = \frac{s + s - c}{s} = \frac{2s - c}{s}$.Since $2s = a + b + c$, we have $\frac{a + b + c - c}{s} = \frac{a + b}{s}$.Comparing this with $\frac{k}{s}$, we get $k = a + b$.

In $\triangle ABC$ with usual notations, if $1 + \tan(\frac{A}{2}) \tan(\frac{B}{2}) = \frac{k}{s}$ (where $s$ is the semi-perimeter), then the value of $k$ is...

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