What Is the Value of a+b If (a+√(a²+1))(b+√(b²+1))=1?

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    2017
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SUMMARY

The equation \((a+\sqrt{a^2+1})(b+\sqrt{b^2+1})=1\) leads to the conclusion that \(a + b = 0\). This result is derived from the properties of hyperbolic functions, specifically recognizing that \(a\) and \(b\) can be expressed in terms of hyperbolic sine and cosine. The correct solutions were provided by forum members kaliprasad, lfdahl, and Theia, who successfully demonstrated the relationship between \(a\) and \(b\) through algebraic manipulation.

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anemone
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Here is this week's POTW:

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Suppose that for two real numbers $a$ and $b$ the following equality is true:

$(a+\sqrt{a^2+1})(b+\sqrt{b^2+1})=1$.

Find the value of $a+b$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution::)
1. kaliprasad
2. lfdahl
3. Theia

Solution from Theia:
If the product of two numbers is equal to one, they have to be reciprocals. So if

$$a + \sqrt{a^2 +1} = u \qquad \Rightarrow \qquad a = \frac{u^2 - 1}{2u}$$

then

$$b + \sqrt{b^2 +1} = \frac{1}{u} \qquad \Rightarrow \qquad b = \frac{u^{-2} - 1}{2u^{-1}}$$.

Now, by direct simplification one obtains

$$a + b = \frac{u^2 - 1}{2u} + \frac{u}{2} \cdot \frac{1 - u^2}{u^2} = \frac{u^2 - 1}{2u} + \frac{1 - u^2}{2u} = 0$$.
 

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