if ((√(1+x^2 ))+x)((√(1+y^2 ))+y)=1 with x,y ∈R, find (x+y)^2 =?


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Question Number 189022 by mr W last updated on 10/Mar/23

$i f (\sqrt{1 + x^{2}} + x) (\sqrt{1 + y^{2}} + y) = 1$ $w i t h x, y \in R, f i n d {(x + y)}^{2} = ?$
	Answered by Frix last updated on 11/Mar/23

	$Obviously y = - x \Rightarrow answer is 0$ $[\sqrt{x^{2} + 1} + x = \frac{1}{\sqrt{x^{2} + 1} - x}]$
	Commented by mr W last updated on 11/Mar/23

	$c a n w e b e s u r e t h a t n o o t h e r s o l u t i o n s$ $e x i s t ?$
	Answered by Frix last updated on 11/Mar/23

	$\sqrt{x^{2} + 1} + x = \frac{1}{\sqrt{y^{2} + 1} + y}$ $\sqrt{x^{2} + 1} + x = \sqrt{y^{2} + 1} - y$ $x + y = \sqrt{y^{2} + 1} - \sqrt{x^{2} + 1}$ $x^{2} + 2 x y + y^{2} = y^{2} + 1 + x^{2} + 1 - 2 \sqrt{x^{2} + 1} \sqrt{y^{2} + 1}$ $x y = 1 - \sqrt{x^{2} + 1} \sqrt{y^{2} + 1}$ $1 - x y = \sqrt{x^{2} + 1} \sqrt{y^{2} + 1}$ $1 - 2 x y + x^{2} y^{2} = x^{2} y^{2} + x^{2} + y^{2} + 1$ $- 2 x y = x^{2} + y^{2}$ ${(x + y)}^{2} = 0$
	Commented by mr W last updated on 11/Mar/23

	$n i c e s o l u t i o n, t h a n k s!$
	Answered by MJS_new last updated on 11/Mar/23

	$z = \sinh u \land y = \sinh v$ $e^{u} e^{v} = 1$ $e^{u + v} = 1$ $u + v = 0$ $v = - u$ $\sinh (- u) = - \sinh u$ $y = - x$
	Commented by mr W last updated on 11/Mar/23

	$n i c e s o l u t i o n, t h a n k s!$
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