In any triangle ABC,frac{tan frac{A}{2} - tan | vedclass

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(B) 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)}}$.Substituting these in

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$\frac{a - b}{c}$

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(B) 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)}}$.Substituting these into the expression:$\frac{	an \frac{A}{2} - 	an \frac{B}{2}}{	an \frac{A}{2} + 	an \frac{B}{2}} = \frac{\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} - \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}{\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} + \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}$$= \frac{\frac{\sqrt{s-c}}{\sqrt{s}} \left( \frac{s-b}{\sqrt{(s-a)(s-b)}} - \frac{s-a}{\sqrt{(s-a)(s-b)}} ight)}{\frac{\sqrt{s-c}}{\sqrt{s}} \left( \frac{s-b}{\sqrt{(s-a)(s-b)}} + \frac{s-a}{\sqrt{(s-a)(s-b)}} ight)}$$= \frac{(s-b) - (s-a)}{(s-b) + (s-a)} = \frac{a-b}{2s-a-b}$.Since $2s = a+b+c$,we have $2s-a-b = c$.Therefore,the expression equals $\frac{a-b}{c}$.

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

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