Question

In triangle ABC, prove the following:

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

Answer 1

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ ...(1)

We need to prove:
$\frac{c}{a-b}=\frac{\tan \left({\displaystyle \frac{A}{2}}\right)+\tan \left({\displaystyle \frac{B}{2}}\right)}{\tan \left({\displaystyle \frac{A}{2}}\right)-\tan \left({\displaystyle \frac{B}{2}}\right)}$

Consider

$$\begin{gathered} \mathrm{LHS}=\frac{c}{a-b} \\ =\frac{k\sin C}{k\left(\sin A-\sin B\right)}\left(\mathrm{using}\left(1\right)\right) \\ =\frac{2\sin {\displaystyle \frac{C}{2}}\cos {\displaystyle \frac{C}{2}}}{2\sin \left({\displaystyle \frac{A-B}{2}}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)} \\ =\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }-\left(\mathrm{A}+\mathrm{B}\right)}{2}}\right)\cos {\displaystyle \frac{C}{2}}}{\sin \left({\displaystyle \frac{A-B}{2}}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)}\left(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi }\right) \\ =\frac{\cos {\displaystyle \frac{C}{2}}\cos \left({\displaystyle \frac{A+B}{2}}\right)}{\sin \left({\displaystyle \frac{A-B}{2}}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)} \\ =\frac{\cos {\displaystyle \frac{C}{2}}}{\sin \left({\displaystyle \frac{A-B}{2}}\right)}...\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{RHS}=\frac{\tan {\displaystyle \frac{A}{2}}+\tan {\displaystyle \frac{B}{2}}}{\tan {\displaystyle \frac{A}{2}}-\tan {\displaystyle \frac{B}{2}}} \\ =\frac{{\displaystyle \frac{\sin {\displaystyle \frac{A}{2}}}{\cos {\displaystyle \frac{A}{2}}}}+{\displaystyle \frac{\sin {\displaystyle \frac{B}{2}}}{\cos {\displaystyle \frac{B}{2}}}}}{{\displaystyle \frac{\sin {\displaystyle \frac{A}{2}}}{\cos {\displaystyle \frac{A}{2}}}}-{\displaystyle \frac{\sin {\displaystyle \frac{B}{2}}}{\cos {\displaystyle \frac{B}{2}}}}} \\ =\frac{\sin {\displaystyle \frac{A}{2}}\cos {\displaystyle \frac{B}{2}}+\sin {\displaystyle \frac{B}{2}}\cos {\displaystyle \frac{A}{2}}}{\sin {\displaystyle \frac{A}{2}}\cos {\displaystyle \frac{B}{2}}-\sin {\displaystyle \frac{B}{2}}\cos {\displaystyle \frac{A}{2}}} \\ =\frac{\sin \left({\displaystyle \frac{A+B}{2}}\right)}{\sin \left({\displaystyle \frac{A-B}{2}}\right)} \\ =\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }-\mathrm{C}}{2}}\right)}{\sin \left({\displaystyle \frac{A-B}{2}}\right)} \\ =\frac{\cos {\displaystyle \frac{C}{2}}}{\sin \left({\displaystyle \frac{A-B}{2}}\right)} \\ =\mathrm{LHS}\left(\mathrm{from}\left(2\right)\right) \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$