Question

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

Answer 1

$\tan \frac{A}{2}+\tan \frac{B}{2}=\frac{c}{\left(a-b\right)}$

R.H.S

$\tan \frac{A}{2}+\tan \frac{B}{2}\Rightarrow \sqrt{\frac{\left(S-b\right)\left(S-c\right)}{S\left(S-a\right)}}+\sqrt{\frac{\left(S-a\right)\left(S-c\right)}{S\left(S-b\right)}}\Rightarrow \sqrt{\frac{\left(S-b\right)\left(S-c\right)}{S\left(S-a\right)}}-\sqrt{\frac{\left(S-a\right)(S-c}{S\left(S-b\right)}}\Rightarrow \frac{\left(S-b\right)\sqrt{\left(S-c\right)}+\left(S-a\right)R\left(S-c\right)}{\left(S-b\right)\sqrt{\left(S-c\right)}-\left(S-a\right)R\left(S-c\right)}\Rightarrow \frac{\sqrt{\left(S-c\right)}\left\{S-b+S-a\right\}}{\sqrt{\left(S-c\right)}\left\{S-b-S+a\right\}}=\frac{\left\{2S-\left(a+b\right)\right\}}{\left(a-b\right)}=\frac{\left(a+b+c-a-b\right)}{a-b}\Rightarrow \frac{c}{\left(a-b\right)}$

Hence, proved.