Question

Prove that $\frac{c}{a+b}=\frac{1-\tan \frac{1}{2}A\tan \frac{1}{2}B}{1+\tan \frac{1}{2}A\tan \frac{1}{2}B}$

Answer 1

$\frac{1-\tan \frac{A}{2}\tan \frac{B}{2}}{1+\tan \frac{A}{2}\tan \frac{B}{2}}$

we know that

$tan\frac{A}{2}=\frac{(s-b)(s-c)}{\Delta }$, $tan\frac{B}{2}=\frac{(s-a)(s-c)}{\Delta }$

putting above values

$=\frac{1-\frac{(s-b)(s-c)(s-a)(s-c)}{{△}^2}}{1+\frac{(s-b)(s-c)(s-a)(s-c)}{{△}^2}}$
$=\frac{{△}^2-(s-b)(s-a)(s-c{)}^2}{{△}^2+(s-b)(s-a)(s-c{)}^2}$
Now using heron's formula
$△=\sqrt{s(s-a)(s-b)(s-c)}$
Substituting in the above expression we get,
$=\frac{s(s-a)(s-b)(s-c)-(s-b)(s-a)(s-c{)}^2}{s(s-a)(s-b)(s-c)+(s-b)(s-a)(s-c{)}^2}$
$=\frac{s-(s-c)}{s+(s-c)}$
$=\frac{c}{2s-c}$
$=\frac{c}{a+b}$.