In a triangle ABC,if frac{r}{r_1} = frac{1}{2 | vedclass

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(B) We know that $r = (s-a) 	an \frac{A}{2} = (s-b) 	an \frac{B}{2} = (s-c) 	an \frac{C}{2}$ and $r_1 = s 	an \frac{A}{2}$.Given $\frac{r}{r_1} =

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$2$

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(B) We know that $r = (s-a) 	an \frac{A}{2} = (s-b) 	an \frac{B}{2} = (s-c) 	an \frac{C}{2}$ and $r_1 = s 	an \frac{A}{2}$.Given $\frac{r}{r_1} = \frac{1}{2}$,we have $\frac{(s-a) 	an \frac{A}{2}}{s 	an \frac{A}{2}} = \frac{1}{2}$,which implies $\frac{s-a}{s} = \frac{1}{2}$,so $2s - 2a = s$,or $s = 2a$.Since $2s = a+b+c$,we have $a+b+c = 4a$,which means $b+c = 3a$.We use the identity $	an \frac{B}{2} 	an \frac{C}{2} + 	an \frac{C}{2} 	an \frac{A}{2} + 	an \frac{A}{2} 	an \frac{B}{2} = 1$.This can be written as $	an \frac{A}{2} (	an \frac{B}{2} + 	an \frac{C}{2}) = 1 - 	an \frac{B}{2} 	an \frac{C}{2}$.Using the formula $	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)}}$,we get $	an \frac{B}{2} 	an \frac{C}{2} = \frac{s-a}{s}$.Since $s = 2a$,$\frac{s-a}{s} = \frac{2a-a}{2a} = \frac{1}{2}$.Thus,$	an \frac{A}{2} (	an \frac{B}{2} + 	an \frac{C}{2}) = 1 - \frac{1}{2} = \frac{1}{2}$.Therefore,$4 	an \frac{A}{2} (	an \frac{B}{2} + 	an \frac{C}{2}) = 4 	imes \frac{1}{2} = 2$.

In a triangle $ABC$,if $\frac{r}{r_1} = \frac{1}{2}$,then $4 \tan \frac{A}{2} \left( \tan \frac{B}{2} + \tan \frac{C}{2} \right) = $

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