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

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

Title	proof of tangents law
Canonical name	ProofOfTangentsLaw
Date of creation	2013-03-22 13:11:04
Last modified on	2013-03-22 13:11:04
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	6
Author	CWoo (3771)
Entry type	Proof
Classification	msc 51-00