Answer:
Option C
Explanation:
We have , in $\triangle ABC$
$(\tan \frac{A}{2})(\tan \frac{B}{2})=\frac{3}{4}$
$\Rightarrow \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}=\frac{3}{4}$
$\Rightarrow \sqrt{\frac{(s-b)(s-c)(s-a)(s-c)}{s(s-a).s(s-b)}}=\frac{3}{4}$
$\Rightarrow\frac{(s-c)}{s}=\frac{3}{4}\Rightarrow\frac{\frac{a+b+c}{2}-c}{\frac{a+b+c}{2}}=\frac{3}{4}$
$\Rightarrow$ $ \frac{a+b-c}{a+b+c}=\frac{3}{4}$
$\Rightarrow $ $4a+4b-4c=3a+3b+3c$
$\Rightarrow $ a+b= 7c