If $$ x , y $$ are real numbers such that $$ \big( x+\sqrt{1+x^2}\big) \big( y+\sqrt{1+y^2} \big) =1 $$ , find the value of $$ (x+y)^2 $$

Question

If  $$ x  , y $$ are real numbers such that $$ \big(  x+\sqrt{1+x^2}\big)  \big(  y+\sqrt{1+y^2}  \big)  =1 $$ , find the value of  $$ (x+y)^2 $$

Uniteasy Admin · Accepted Answer

Let 
$$ n = x+\sqrt{1+x^2} $$
The asking expression is simplified as
$$  n(y+\sqrt{1+y^2})=1 $$
$$ y+\sqrt{1+y^2} = \dfrac{1}{n} $$
$$ \sqrt{1+y^2} = \dfrac{1}{n}-y $$
$$ \Big( \sqrt{1+y^2}\Big) ^2 = \Big( \dfrac{1}{n}-y\Big) ^2 $$
$$ 1+y^2 = \dfrac{1}{n^2}- \dfrac{2y}{n}+y^2 $$
$$ \dfrac{1}{n^2}- \dfrac{2y}{n}=1 $$
$$ -2y = n-\dfrac{1}{n} $$
$$ y = \dfrac{1}{2}(\dfrac{1}{n}-n) $$n
$$ \dfrac{1}{n}-n $$
$$ =\dfrac{1}{ x+\sqrt{1+x^2}}-( x+\sqrt{1+x^2}) $$
$$ =\dfrac{ x-\sqrt{1+x^2}}{( x+\sqrt{1+x^2})( x-\sqrt{1+x^2})} -( x+\sqrt{1+x^2}) $$
$$ =\dfrac{ x-\sqrt{1+x^2}}{-1} -( x+\sqrt{1+x^2}) $$
$$ =-2x $$
$$ 	herefore y = -x $$
$$ (x+y)^2=0 $$

Question

Answer