Proving Solution of Triangles with Tan Function

[DrunkenOldFool]

In a triangle $ABC$, the sides opposite to vertices $A,B,C$ are $a,b,c$ respectively. I have to prove

$$ \frac{\tan\left( \frac{A}{2}\right)}{(a-b)(a-c)}+\frac{\tan\left( \frac{B}{2}\right)}{(b-c)(b-a)}+\frac{\tan\left( \frac{C}{2}\right)}{(c-a)(c-b)} = \frac{1}{\Delta}$$

$\Delta$ denotes the area of the triangle.

 

[MarkFL]

I would consider the half-angle identity for tangent:

$\displaystyle \tan\left(\frac{\theta}{2} \right)=\frac{1-\cos(\theta)}{\sin(\theta)}$

and the formula:

$\displaystyle \Delta=\frac{1}{2}bc\sin(A)$

and the law of sines:

$\displaystyle \frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$

to write the expression as:

$\displaystyle \frac{1-\cos(A)}{\sin(A)(a-b)(a-c)}+\frac{a(1-\cos(B))}{b\sin(A)(b-c)(b-a)}+\frac{a(1-\cos(C))}{c\sin(A)(c-a)(c-b)}=\frac{2}{bc\sin(A)}$

Multiply through by $\displaystyle bc\sin(A)$

$\displaystyle \frac{bc(1-\cos(A))}{(a-b)(a-c)}+\frac{ac(1-\cos(B))}{(b-c)(b-a)}+\frac{ab(1-\cos(C))}{(c-a)(c-b)}=2$

Multiply through by $\displaystyle (a-b)(a-c)(b-c)$:

$\displaystyle bc(b-c)(1-\cos(A))+ac(c-a)(1-\cos(B))+ab(a-b)(1-\cos(C))=2(a-b)(a-c)(b-c)$

Using the law of cosines, we may write:

$\displaystyle 1-\cos(A)=1-\frac{b^2+c^2-a^2}{2bc}=\frac{a^2-(b-c)^2}{2bc}$

$\displaystyle 1-\cos(B)=1-\frac{a^2+c^2-b^2}{2ac}=\frac{b^2-(a-c)^2}{2ac}$

$\displaystyle 1-\cos(C)=1-\frac{a^2+b^2-c^2}{2ab}=\frac{c^2-(a-b)^2}{2ab}$

and we have:

$\displaystyle (b-c)(a^2-(b-c)^2)+(c-a)(b^2-(a-c)^2)+(a-b)(c^2-(a-b)^2)=4(a-b)(a-c)(b-c)$

I shall leave you now to verify (as I have), that this is an identity.(Mmm)