In a triangle ABC,if 4a = b + c,then tan frac | vedclass

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(B) We know that $	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)}}$.Multiplying these,we

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$\frac{3}{5}$

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(B) We know that $	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)}}$.Multiplying these,we get $	an \frac{B}{2} 	an \frac{C}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)} \cdot \frac{(s-a)(s-b)}{s(s-c)}} = \sqrt{\frac{(s-a)^2}{s^2}} = \frac{s-a}{s}$.Substituting $s = \frac{a+b+c}{2}$,we have $\frac{s-a}{s} = \frac{\frac{a+b+c}{2} - a}{\frac{a+b+c}{2}} = \frac{b+c-a}{b+c+a}$.Given $b+c = 4a$,we substitute this into the expression:$\frac{4a-a}{4a+a} = \frac{3a}{5a} = \frac{3}{5}$.

In a $\triangle ABC$,if $4a = b + c$,then $\tan \frac{B}{2} \tan \frac{C}{2} =$

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