Question

In $\Delta ABC$ if $x=\tan \left(\frac{B-C}{2}\right)\tan \frac{A}{2},y=\tan \left(\frac{C-A}{2}\right)\tan \frac{B}{2}$ and $z=\tan \left(\frac{A-B}{2}\right)\tan \frac{C}{2}$, then $(x+y+z)$ is equal to

• A. x y z
• B. -x y z
• C. 2 x y z
• D. $\frac{1}{2}xyz$

Answer 1

Answer: (B)

Given, $x=\tan \left(\frac{B-C}{2}\right)\tan \frac{A}{2}$
$y=\tan \left(\frac{C-A}{2}\right)\tan \frac{B}{2}$
and $z=\tan \left(\frac{A-B}{2}\right)\tan \frac{C}{2}$
$\Rightarrow x=\frac{b-c}{b+c}$
$\left[\because \tan \frac{B-C}{2}=\frac{b-c}{b+c}\cot \frac{A}{2}\right]$
$y=\frac{c-a}{c+a}$ and $z=\frac{a-b}{a+b}$
Now, $\frac{1+x}{1-x}=\frac{b+c+b-c}{b+c-b+c}$[By componendo and dividendo]
$\Rightarrow \frac{1+x}{1-x}=\frac{b}{c}$
Similarly, $\frac{1+y}{1-y}=\frac{c}{a}$ and $\frac{1+z}{1-z}=\frac{a}{b}$
$\therefore \left(\frac{1+x}{1-x}\right)\left(\frac{1+y}{1-y}\right)\left(\frac{1+z}{1-z}\right)=\frac{b}{c}\times \frac{c}{a}\times \frac{a}{b}=1$
$\Rightarrow (1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)$
$\Rightarrow 1+x+y+z+xy+yz+zx+xyz=1-(x+y+z)+xy+yz+zx-xyz$
$\Rightarrow x+y+z=-xyz$