Question

if x =a-b/a+b, y=b-c/b+c, z=c-a/c+a, then find the value of (1+x)(1+y)1+z)/1-x)(1-y)(1-z).​

Answer 1

give me best ans and thanks plzzzzz

Answer 2

Answer:

Answer would be -1

Step-by-step explanation:

Given,

$x=\frac{a-b}{a+b}, y =\frac{b-c}{b+c}, z=\frac{c-a}{c+a}----(1)$

$\implies 1+x =1+\frac{a-b}{a+b}, 1+y=1+\frac{b-c}{b+c}, 1+z=1+\frac{c-a}{c+a}$

$\implies 1+x =\frac{a+b+a-b}{a+b}, 1+y=\frac{b+c+b-c}{b+c}, 1+z=\frac{c+a+c-a}{c+a}$

$\implies 1+x =\frac{2a}{a+b}, 1+y=\frac{2b}{b+c}, 1+z=\frac{2c}{c+a}---(2)$

Again from equation (1),

$\implies 1-x =1-\frac{a-b}{a+b}, 1-y=1-\frac{b-c}{b+c}, 1-z=1-\frac{c-a}{c+a}$

$\implies 1-x =\frac{a+b-a+b}{a+b}, 1-y=\frac{b+c-b+c}{b+c}, 1-z=\frac{c+a-c+a}{c+a}$

$\implies 1-x =\frac{2b}{a+b}, 1-y=\frac{2c}{b+c}, 1-z=\frac{2a}{c+a}---(3)$

From equation (2) and (3),

$\frac{1+x}{1-x}=\frac{a}{b}, \frac{1+y}{1-y}=\frac{b}{c}, \frac{1+z}{1-z}=\frac{c}{a}$

$\implies \frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}=\frac{abc}{abc}=1$

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