Given $\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$, prove $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$.

Question

Let $x$ and $y$ be real numbers such that $$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$$ Prove that $$\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$$

Answer 1

We'll replace $y$ an $-y$.

Thus, the given it's $$\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-y\right)=1$$ and we need to prove that: $$\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)=1.$$ Now, the condition gives $$\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-y\right)=\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)$$ or $$\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-y\right)-\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)+$$ $$+\left(\sqrt{y^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)-\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)=0.$$ or $$(x-y)\left(\sqrt{y^2+1}+x\right)-\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}-\sqrt{y^2+1}\right)=0$$ or $$(x-y)\left(\sqrt{y^2+1}+x-\frac{(x+y)\left(\sqrt{x^2+1}-x\right)}{\sqrt{x^2+1}+\sqrt{y^2+1}}\right)=0,$$ which gives $x=y$ because the second factor it's $$\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)+(x+y)^2+1>0.$$ Id est, $$\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}-y\right)=\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)=1$$ and we are done!

Answer 2

Hint :

By the hypothesis of the problem, do some algebra and show that :

$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1 \Leftrightarrow \dots \Leftrightarrow y = -x$$

Now, substitute $y=-x$ on the expression $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right)$ and see what happens.

Answer 3

Write $x=\tan a$ and $y = \tan b$ for some (angles) $a,b$. They exsist since $\tan $ is surjective function. Pluging in starting equation we get:

$$ {\sin a+1\over \cos a}\cdot {\sin b+1\over \cos b}=1$$ and after rearranging we get $$\sin(a)+ \sin(b) = \cos (a+b) - \cos 0$$

which is equivalent to $$2\sin{a+b\over 2}\cos{a-b\over 2} = -2\sin {a+b\over 2}\sin{a+b\over 2}$$

Case 1. $\sin {a+b\over 2}=0$ then $a+b = 2\pi k$ so $$x=\tan a = \tan (2\pi k-b) = -\tan b = -y$$

Case 2. $\sin {a+b\over 2}\ne 0$, then $$\cos {a-b\over 2}+\sin{a+b\over 2}=0$$ Factorising this we get:

$$2\cos ({\pi\over 4}-{b\over 2})\cdot \cos({a\over 2}-{\pi\over 4})=0$$

Now we have to choises again.

1.st: $${\pi\over 4}-{b\over 2} = {\pi\over 2}+\pi k\implies b =-{\pi\over 2}+2\pi k $$ so $y$ does not exist.

2.nd case... we get $x$ does not exsist.

So $x=-y$ and thus second expresins is also $1$.

Answer 4

Hint:

Like greedoid, WLOG let $x=\cot2A,y=\cot2B,0<2A,2B<\pi$ (Reference )

$$1=(x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=\cot A\cot B$$

$\implies\tan A=\cot B$

$x=\cot2A=\dfrac{1-\tan^2A}{2\tan A}=\dfrac{1-\cot^2B}{2\cot B}=\cdots=-\cot2B=-y$