proof of tangents law

To prove that

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

we start with the sines law, which says that

\frac{a}{\sin(A)}=\frac{b}{\sin(B)}.

This implies that

a\sin(B)=b\sin(A)

We can write $\sin(A)$ as

\sin(A)=\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})\sin(\frac{A% -B}{2}).

and $\sin(B)$ as

\sin(B)=\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})-\cos(\frac{A+B}{2})\sin(\frac{A% -B}{2}).

Therefore, we have

a(\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})-\cos(\frac{A+B}{2})\sin(\frac{A-B}{2}% ))=b(\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})\sin(\frac{A-B}% {2}))

Dividing both sides by $\cos(\frac{A-B}{2})\cos(\frac{A+B}{2}),$ we have,

a(\tan(\frac{A+B}{2})-\tan(\frac{A-B}{2}))=b(\tan(\frac{A+B}{2})+\tan(\frac{A-% B}{2}))

This gives us

\frac{a}{b}=\frac{\tan(\frac{A+B}{2})+\tan(\frac{A-B}{2})}{\tan(\frac{A+B}{2})% -\tan(\frac{A-B}{2})}

Hence we find that

\frac{a-b}{a+b}=\frac{\displaystyle{\frac{a}{b}}-1}{\displaystyle{\frac{a}{b}}% +1}=\frac{\tan(\frac{A-B}{2})}{\tan(\frac{A+B}{2})}.