Question

In $△ABC$, let $x=\tan \frac{B-C}{2}\tan \frac{A}{2},y=\tan \frac{C-A}{2}\tan \frac{B}{2},z=\tan \frac{A-B}{2}\tan \frac{C}{2}$
Then $x+y+z=K\left(xyz\right)$, where the value of $K$ is

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$

Answer 1

The correct option is D $-1$

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

Similarly,$y=\left(\frac{c-a}{c-b}\right)z=\left(\frac{a-b}{a+b}\right)x+y9z=k\left(xyz\right)\frac{b-c}{b+c}+\frac{c-a}{c-b}+\frac{a-b}{a+b}=k\left(xyz\right)\Rightarrow \frac{(b-c)(a+c)(a+b)+(c-a)(b+c)(a+b)+(a-b)(b+c)(a+c)}{(b+c)(c+a)(a+b)}=k\left(xyz\right)$

$\Rightarrow \frac{(a+b)(ab+bc-ac-c^2+bc+c^2-ab-ac)+(a-b)(b+c)(a+c)}{(b+c)(c+a)(a+b)}=k\left(xyz\right)$

$\Rightarrow \frac{2(a+b)(bc-a{c}^2)+(a-b)(b+c)(a+c)}{(b+c)(c+a)(a+b)}=k\left(xyz\right)$

$\Rightarrow \frac{(b-a)(2ac-2bc-(b+c)(a+c))}{(b+c)(c+a)(a+b)}=k\left(xyz\right)$

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

$\Rightarrow -(a-b)(ac+bc-ab-c^2)=k(b-c)(c-a)(a-b)\Rightarrow (ab-bc-ac+c^2)=k(bc-ab+ac-c^2)k=-1$