(a-b)/(a+b)=[tan{(A-B)}/2]/[tan{(A+B)}/2]
Hello matsci0000You don't say so, but I presume that $\displaystyle a, b, A, B$ are sides and angles in a triangle. In which case, using the Sine Rule:
$\displaystyle a=\frac{b\sin A}{\sin B}$
$\displaystyle \Rightarrow \frac{a-b}{a+b}= \frac{\frac{b\sin A}{\sin B}-b}{\frac{b\sin A}{\sin B}+b}=\frac{\sin A -\sin B}{\sin A + \sin B}$
Now use the identities $\displaystyle \sin A +\sin B=2\sin \tfrac12 (A+B)\cos \tfrac12(A-B)$ and $\displaystyle \sin A -\sin B=2\sin\tfrac12(A-B)\cos\tfrac12(A+B)$:
$\displaystyle \Rightarrow \frac{a-b}{a+b}= \frac{2\sin\tfrac12(A-B)\cos\tfrac12(A+B)}{2\sin\tfrac12(A+B)\cos\tfrac1 2(A-B)}$
$\displaystyle = \frac{\tan\tfrac12(A-B)}{\tan\tfrac12(A+B)}$
Grandad