Question

In a triangle $ABC$ we define
$x=tan\frac{B-C}{2}tan\frac{A}{2},y=tan\frac{C-A}{2}tan\frac{B}{2}$ and $z=tan\frac{AB}{2}tan\frac{C}{2}$
Then the value of $x+y+z$ (in terms of $x,y,z$) is

• A. xyz
• B. -xyz
• C. 2xyz
• D. none

Answer 1

The correct option is B

-xyz

$tan\frac{B-C}{2}=\frac{b-c}{b+c}cot\frac{A}{2},\text{Napier's analogy}\therefore \frac{x}{1}=tan\frac{B-C}{2}tan\frac{A}{2}=\frac{b-c}{b+c}\Rightarrow xb+xc=b-c\Rightarrow \frac{c(x+1)}{b(1-x)}\therefore \frac{1+x}{1-x}=\frac{b}{c}\text{Similarly}\frac{1+y}{1-y}=\frac{c}{a}\text{and}\frac{1+z}{1-z}=\frac{a}{b}\therefore \frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}=\frac{b}{c}.\frac{c}{a}.\frac{a}{b}=1\therefore \frac{(1+x)(1+y+z+yz)}{(1-x)(1-y-z+yz)}=1\Rightarrow 1+y+z+yz+x+xy+xz+xyz=1-y-z+yz-x+xy+xz-xyz\Rightarrow 2(x+y+z)=-2xyz\therefore (x+y+z)=-xyz$