Vrftsjtryk

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
$$left(x + sqrt{1+x^2}right)left(y + sqrt{1+y^2}right) = 1$$

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

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

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.

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:

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

which is equivalent to $$2sin{a+bover 2}cos{a-bover 2} = -2sin {a+bover 2}sin{a+bover 2}$$

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

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

$$2cos ({piover 4}-{bover 2})cdot cos({aover 2}-{piover 4})=0$$

Now we have to choises again.

1.st: $${piover 4}-{bover 2} = {piover 2}+pi kimplies b =-{piover 2}+2pi k $$
so $y$ does not exist.

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

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

Hint:

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

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

$impliestan A=cot B$

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