Question

solv this for me....... ​

Answer 1

Solution :

Given :

          x = $\displaystyle\frac{a-b}{a+b}$

          y = $\displaystyle\frac{b-c}{b+c}$

          z = $\displaystyle\frac{c-a}{c+a}$

Now, $\displaystyle\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}$

$\displaystyle=\frac{(1+\frac{a-b}{a+b})(1+\frac{b-c}{b+c})(1+\frac{c-a}{c+a})}{(1-\frac{a-b}{a+b})(1-\frac{b-c}{b+c})(1-\frac{c-a}{c+a})}$

$\displaystyle=\frac{\frac{a+b+a-b}{a+b}\times\frac{b+c+b-c}{b+c}\times\frac{c+a+c-a}{c+a}}{\frac{a+b-a+b}{a+b}\times\frac{b+c-b+c}{b+c}\times\frac{c+a-c+a}{c+a}}$

$\displaystyle=\frac{2a\times 2b\times 2c}{2b\times 2c\times 2a}$

$\displaystyle=1$

$\to \boxed{\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}=1}$

Answer 2

hii mate check your attachment plse!!