Question

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

• A. $\frac{a-b}{a+b}$
• B. $\frac{a-b}{c}$
• C. $\frac{a-b}{a+b+c}$
• D. $\frac{c}{a+b}$

Answer 1

The correct option is B $\frac{a-b}{c}$

Using formula of incentre; $r=(s-a)tan\frac{A}{2};$
$\Rightarrow tan\frac{A}{2}=\frac{r}{s-a}$
Similarly; $tan\frac{B}{2}=\frac{r}{s-b}$
$\frac{tan\frac{A}{2}-tan\frac{B}{2}}{tan\frac{A}{2}+tan\frac{B}{2}}=\frac{\frac{r}{s-a}-\frac{r}{s-b}}{\frac{r}{s-a}+\frac{r}{s-b}}$
$=\frac{(s-b)-(s-a)}{(s-b)+(s-a)}$
$=\frac{a-b}{2s-a-b}$
$=\frac{a-b}{c}$ $[\text{Since}s=\frac{a+b+c}{2}]$