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})}.