Answer 1

\[\text{ Assume }\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\] 

Consider the LHS of the equation 

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

\[LHS = \frac{a - b}{a + b}\]
\[ = \frac{k\left( \sin A - \sin B \right)}{k\left( \sin A + \sin B \right)}\]

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

\[\therefore 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)} $$\]
\[ = \frac{\tan\left( \frac{A - B}{2} \right)}{\tan\left( \frac{A + B}{2} \right)} = RHS \]
\[\text{ Hence proved } .\]