# In triangle ABC, prove the following:

Question:

In triangle ABC, prove the following:

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

Solution:

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

Consider the LHS of the equation $\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}$.

$\mathrm{LHS}=\frac{a-b}{a+b}$

$=\frac{k(\sin A-\sin B)}{k(\sin A+\sin B)}$

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

$\therefore \mathrm{LHS}=\frac{2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}$

$=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}=\mathrm{RHS}$

Hence proved.
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