In triangle ABC,if x=tan left(frac{B-C}{2}rig | vedclass

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(B) In $	riangle ABC$,we use the Napier's analogy: $	an \left(\frac{B-C}{2}ight) = \frac{b-c}{b+c} \cot \frac{A}{2}$.Given $x = 	an \left(\frac{B

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$-xyz$

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(B) In $	riangle ABC$,we use the Napier's analogy: $	an \left(\frac{B-C}{2}ight) = \frac{b-c}{b+c} \cot \frac{A}{2}$.Given $x = 	an \left(\frac{B-C}{2}ight) 	an \frac{A}{2}$,substituting the analogy gives $x = \frac{b-c}{b+c} \cot \frac{A}{2} 	an \frac{A}{2} = \frac{b-c}{b+c}$.Similarly,$y = \frac{c-a}{c+a}$ and $z = \frac{a-b}{a+b}$.Now,consider $\frac{1+x}{1-x} = \frac{1 + \frac{b-c}{b+c}}{1 - \frac{b-c}{b+c}} = \frac{b+c+b-c}{b+c-b+c} = \frac{2b}{2c} = \frac{b}{c}$.Similarly,$\frac{1+y}{1-y} = \frac{c}{a}$ and $\frac{1+z}{1-z} = \frac{a}{b}$.Multiplying these,we get $\left(\frac{1+x}{1-x}ight) \left(\frac{1+y}{1-y}ight) \left(\frac{1+z}{1-z}ight) = \frac{b}{c} \cdot \frac{c}{a} \cdot \frac{a}{b} = 1$.This implies $(1+x)(1+y)(1+z) = (1-x)(1-y)(1-z)$.Expanding both sides: $1 + (x+y+z) + (xy+yz+zx) + xyz = 1 - (x+y+z) + (xy+yz+zx) - xyz$.Simplifying,we get $2(x+y+z) = -2xyz$,which means $x+y+z = -xyz$.

In $\triangle ABC$,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

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