Question

(x+1)/(x-1)=a/b, (1-y)/(1+y)=b/a then (x-y)/(1+xy)=?​

Answer 1

Answer:

Ab is the answer

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Just Find out Value of X then substitute with value of Y

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Answer 2

Given,

$\frac{x+1}{x-1}=\frac{a}{b}\\\implies \frac{x+1}{x-1}+1=\frac{a}{b}+1\\\implies \frac{x+1+x-1}{x-1}=\frac{a+b}{b}\\\implies \frac{2x}{x-1}=\frac{a+b}{b}\\\implies 2bx =(a+b)(x-1)\\\implies 2bx =(a+b)x-(a+b)\\\implies (a+b)x-2bx =a+b\\\implies x(a+b-2b)=a+b\\\implies x=\frac{a+b}{a-b} \:--\: [eqtn.\:1]$

Also,

$\frac{1-y}{1+y}=\frac{b}{a}\\\implies\frac{1-y}{1+y}+1=\frac{b}{a}+1\\\implies\frac{1-y+1+y}{1+y}=\frac{b+a}{a}\\\implies\frac{2}{1+y} =\frac{a+b}{a} \\\implies 2a = (a+b)(1+y)\\\implies \frac{2a}{a+b} = 1+y\\\implies y=\frac{2a}{a+b}-1=\frac{2a-a-b}{a+b}=\frac{a-b}{a+b}\:--\:[eqtn.\:2]$

Now,

$\frac{x-y}{1+xy}=\frac{\frac{a+b}{a-b} -\frac{a-b}{a+b} }{1+\frac{a+b}{a-b} \times\frac{a-b}{a+b} }\\\\=\frac{\frac{(a+b)^2-(a-b)^2}{(a+b)(a-b)} }{1+1} \\\\=\frac{1}{2}\times\frac{(a+b)^2-(a-b)^2}{(a+b)(a-b)}\\=\frac{1}{2}\times\frac{a^2+2ab+b^2-(a^2-2ab+b^2)}{(a+b)(a-b)}\\=\frac{1}{2}\times\frac{a^2+2ab+b^2-a^2+2ab-b^2}{(a+b)(a-b)}\\==\frac{1}{2}\times\frac{4ab}{(a+b)(a-b)}\\\therefore \frac{x-y}{1+xy} =\frac{ab}{(a+b)(a-b)}$