If the real number $$ x $$ , $$ y $$ satisfy the condition
$$ \big( x+\sqrt{1+x^2}\big) \big( y+\sqrt{1+y^2} \big) =1 $$ ,
verify the value of the expression $$ (x+y)^2$$ is equal to 0.

Question

If the real number $$ x $$ , $$ y $$ satisfy the condition  
$$ \big(  x+\sqrt{1+x^2}\big)  \big(  y+\sqrt{1+y^2}  \big)  =1 $$ , 
verify the value of the expression $$ (x+y)^2$$  is equal to 0.

Uniteasy Admin · Accepted Answer

Let $$ n = x+\sqrt{1+x^2} $$n
The condition expression is transformed to ,
$$  n(y+\sqrt{1+y^2})=1 $$
$$ y+\sqrt{1+y^2} = \dfrac{1}{n} $$
$$ \sqrt{1+y^2} = \dfrac{1}{n}-y $$
Square the both sides of the equation
$$ \Big( \sqrt{1+y^2}\Big) ^2 = \Big( \dfrac{1}{n}-y\Big) ^2 $$
$$ 1+y^2 = \dfrac{1}{n^2}- \dfrac{2y}{n}+y^2 $$n
$$ \dfrac{1}{n^2}- \dfrac{2y}{n}=1 $$
$$ -2y = n-\dfrac{1}{n} $$
$$ y = \dfrac{1}{2}(\dfrac{1}{n}-n) $$n
Replace n with (1)
$$ \dfrac{1}{n}-n $$
$$ =\dfrac{1}{ x+\sqrt{1+x^2}}-( x+\sqrt{1+x^2}) $$
$$ =\dfrac{ x-\sqrt{1+x^2}}{( x+\sqrt{1+x^2})( x-\sqrt{1+x^2})} -( x+\sqrt{1+x^2}) $$
$$ =\dfrac{ x-\sqrt{1+x^2}}{-1} -( x+\sqrt{1+x^2}) $$
$$ =-2x $$
$$ 	herefore y = -x $$
$$ (x+y)^2=0 $$

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