Question

If $x\ne 0$ and $x\ne 1$ and if $x$ is replaced by $\frac{1}{x}$ everywhere in the expression ${\left(\frac{x+1}{x-1}\right)}^2$
then the resulting expression is equivalent to

• A. ${\left(\frac{x+1}{1-x}\right)}^2$
• B. ${\left(\frac{x-1}{x+1}\right)}^2$
• C. $\frac{x^2+1}{1-x^2}$
• D. $\frac{x^2-1}{x^2+1}$
• E. $-{\left(\frac{x-1}{x+1}\right)}^2$

Answer 1

The correct option is A ${\left(\frac{x+1}{1-x}\right)}^2$

Given expression is ${\left(\frac{x+1}{x-1}\right)}^2$

Replace $x$ by $\frac{1}{x}$ in the above expressions

So, the expression will be

$\frac{(x+1{)}^2}{(x-1{)}^2}$ $=\frac{{(\frac{1}{x}+1)}^2}{{(\frac{1}{x}-1)}^2}$

After solving above expression, we get

$\frac{(x+1{)}^2}{(1-x{)}^2}={\left(\frac{x+1}{1-x}\right)}^2$

Hence, option A is correct.