Question

In triangle ABC, prove the following:

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

Answer 1

$\mathrm{Assume}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Consider the LHS of the equation $\frac{a-b}{a+b}=\frac{\tan \left({\displaystyle \frac{A-B}{2}}\right)}{\tan \left({\displaystyle \frac{A+B}{2}}\right)}$.
$$\begin{gathered} \mathrm{LHS}=\frac{a-b}{a+b} \\ =\frac{k\left(\sin A-\sin B\right)}{k\left(\sin A+\sin B\right)} \end{gathered}$$

$\because \sin A-\sin B=2\sin \left({\displaystyle \frac{A-B}{2}}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right),\sin A+\sin B=2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)$.

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