Question

$In△ABC,ifx=\tan \left(\frac{B-C}{2}\right)\tan \frac{A}{2},y=\tan \left(\frac{C-A}{2}\right)\tan \frac{B}{2},z\tan \left(\frac{A-B}{2}\right)\tan \frac{C}{2},thenx+y+z$ (in terms of x, y, z only) is

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

Answer 1

The correct option is C $-xyz$

$\text{Solve: Given,}$

$x=\tan \left(\frac{B-C}{2}\right)\tan \frac{A}{2}-\left(i\right)$

$y=\tan \left(\frac{c-A}{2}\right)\tan \frac{B}{2}-\left(ii\right)$

$z=\tan \left(\frac{A-B}{2}\right)\tan \frac{C}{2}-\left(ii\right)$

From solution of triangle we know

that, $\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot \frac{A}{2}$

on putting the value of $\tan \left(\frac{B-C}{2}\right)$ in

eqn. (i) we get,

$x=\frac{b-c}{b+c}\times \cot \frac{A}{2}\times \tan \frac{A}{2}$

$\Rightarrow x=\frac{b-c}{b+c}-\left(iii\right)$

$\text{similary,}y=\frac{c-a}{c+a}-\left(iv\right)$

and $z=\frac{a-b}{a+b}\left(v\right)$

on adding equation (ii), (iv) and (v) we get

$x+y+z=\frac{b-c}{b+c}+\frac{c-a}{c+a}+\frac{a-b}{a+b}$

$\Rightarrow x+y+z=\frac{(b-c)(c+a)(a+b)+(c-a)(b+c)(a+b)+(a-b)(a+c)(b+c)}{(b+c)(c+a)(a+b)}$

$=\frac{(a+b)(bc-c^2+ab-ac+bc-ab+c^2-ac]+(a-b)(a+c)(b+c)}{(b+c)(c+a)(a+b)}$

$=\frac{(a+b)2c(b-a)+(a-b)(a+c)(b+c)}{(b+c)(c+a)(a+b)}$

$=\frac{(a-b)[-2ac-2bc+ab+ac+bc+c^2]}{(a+b)(b+c)(c+a)}$

$=\frac{(a-b)\left[c\right(c-b)-a(c-b\left)\right]}{(a+b)(b+c)(c+a)}$

$=\frac{(a-b)(c-a)(c-b)}{(a+b)(b+c)(c+a)}$

$=\left(\frac{a-b}{a+b}\right)\left(\frac{c-b}{b+c}\right)\left(\frac{c-a}{c+a}\right)$

$=\left(x\right)(-y)\left(z\right)x+y+z=-xyz$