4

I am trying to prove a hyperbolic trigonometric identity and I ran into the following expression:

$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} \quad.$$

This expression is supposed to be equivalent to

$$\sqrt{x^2+1} \quad.$$

I tried to algebraically manipulate the original expression to get the required expression and got to:

$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} = x+\frac{1}{\sqrt{x^2+1} + x} \quad.$$

but I don't know how I can show that

$$x+\frac{1}{\sqrt{x^2+1} + x} = \sqrt{x^2+1}\quad .$$

user4010936
  • 75

3 Answers3

7

What you got is correct . Then, multiply $$\frac{1}{\sqrt{x^2+1}+x}$$ by $$\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x}(=1).$$

mathlove
  • 139,939
1

For $b\ne 0$, $\frac{a}{b}=c\Leftrightarrow bc=a$.

So a strategy would be to try to show that $$(\sqrt{x^2+1}+x)^2+1=2(\sqrt{x^2+1}+x)\cdot \sqrt{x^2+1}$$

paw88789
  • 40,402
1

$$\\an-inovative-solution\\ \\x=tan(a) \Rightarrow \sqrt{x^2+1}=sec(a) \\\frac{(\sqrt{x^2+1}+x)^2+1}{2(\sqrt{x^2+1}+x)}=\\\frac{(sec(a)+tan(a))^2+1}{2(sec(a)+tan(a))}=\\\frac{(sec^2(a)+tan^2(a)+2sec(a)tan(a))+1}{2(sec(a)+tan(a)}=\\\frac{(sec^2(a)+tan^2(a)+2sec(a)tan(a))+1}{2(sec(a)+tan(a))}=\\as-we-know-(sec^2(a)=1+tan^2(a))\\so\\\frac{(sec^2(a)+sec^2(a)+2sec(a)tan(a))}{2(sec(a)+tan(a)}=\\\frac{2sec^2(a)+2sec(a)tan(a)}{2(sec(a)+tan(a)}=\\\frac{2sec^2(a)+2sec(a)tan(a))}{2(sec(a)+tan(a))}=\\\frac{2sec(a)(sec(a)+tan(a)}{2(sec(a)+tan(a))}=\\sec(a)=\sqrt{x^2+1} $$

Khosrotash
  • 24,922