# proof of tangents law

To prove that

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

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

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

 $\frac{a-b}{a+b}=\frac{\displaystyle{\frac{a}{b}}-1}{\displaystyle{\frac{a}{b}}% +1}=\frac{\tan(\frac{A-B}{2})}{\tan(\frac{A+B}{2})}.$
Title proof of tangents law ProofOfTangentsLaw 2013-03-22 13:11:04 2013-03-22 13:11:04 CWoo (3771) CWoo (3771) 6 CWoo (3771) Proof msc 51-00