Question Number 206063 by MATHEMATICSAM last updated on 06/Apr/24
$$\mathrm{If}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\left({y}\:+\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{2}} }\right)\:=\:\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{find}\:\left({x}\:+\:{y}\right)^{\mathrm{2}} . \\ $$
Answered by mr W last updated on 06/Apr/24
$${x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }=−{y}+\sqrt{\mathrm{1}+\left(−{y}\right)^{\mathrm{2}} } \\ $$$${f}\left({x}\right)={x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{is}\:{strictly}\:{increasing}. \\ $$$${if}\:{f}\left({a}\right)={f}\left({b}\right),\:{then}\:{a}={b}. \\ $$$$\Rightarrow{x}=−{y} \\ $$$$\Rightarrow{x}+{y}=\mathrm{0}\:\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$
Answered by Berbere last updated on 06/Apr/24
$${y}={sh}\left({t}\right)={x}={sh}\left({w}\right) \\ $$$$\left({sh}\left({w}\right)+{ch}\left({w}\right)\right)\left({sh}\left({t}\right)+{ch}\left({t}\right)\right)=\mathrm{1} \\ $$$${e}^{{w}} .{e}^{{t}} =\mathrm{1} \\ $$$${w}+{t}=\mathrm{0} \\ $$$${x}+{y}={sh}\left({w}\right)+{sh}\left(−{w}\right)=\mathrm{0} \\ $$