1. Simple algebra

Looks like simple algebra but can't do it.

The numerical vale of the following forms come equal. But I can't convert the 1st form into 2nd :-

1st. $\displaystyle ln {\frac{(\sqrt{1+x^2} - 1)(\sqrt{1+y^2} - 1)}{|xy|}}$

2nd. $\displaystyle -ln {\frac{(\sqrt{1+x^2} + 1)(\sqrt{1+y^2} + 1)}{|xy|}}$

2. Originally Posted by Altair
Looks like simple algebra but can't do it.

The numerical vale of the following forms come equal. But I can't convert the 1st form into 2nd :-

1st. $\displaystyle ln {\frac{(\sqrt{1+x^2} - 1)(\sqrt{1+y^2} - 1)}{|xy|}}$

2nd. $\displaystyle -ln {\frac{(\sqrt{1+x^2} + 1)(\sqrt{1+y^2} + 1)}{|xy|}}$

For this to be true, then:

$\displaystyle \frac{(\sqrt{1+x^2} - 1)(\sqrt{1+y^2} - 1)}{|xy|} = \left[\frac{|xy|}{(\sqrt{1+x^2} + 1)(\sqrt{1+y^2} + 1)}\right]$

$\displaystyle (\sqrt{1+x^2} - 1)(\sqrt{1+y^2} - 1)(\sqrt{1+x^2} + 1)(\sqrt{1+y^2} + 1) = (xy)^2$

$\displaystyle (1+x^2-1)(1+y^2-1) = (xy)^2$

$\displaystyle (xy)^2 = (xy)^2$

3. I want the conversion from 1st form to the 2nd.Not its proof.

4. Originally Posted by Altair
I want the conversion from 1st form to the 2nd.Not its proof.
Showing the path from one form to another is a proof. It is unfortunate that my answer is not good enough for you.

Why did I setup the first line of my problem like so?

This comes from a log property: $\displaystyle log(a^b)=b*log(a)$
In this case, $\displaystyle b=-1$ hence why the 2nd term in my 1st line is a reciprocal of what is being logged in your 2nd equation.