<s>[INST] Show that two expressions are equivalent 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
$$x+\frac{1}{\sqrt{x^2+1} + x} = \sqrt{x^2+1}\quad .$$
 [/INST] $$\\an-inovative-solution\\ \\x=tan(a) \Rightarrow \sqrt{x^2+1}=sec(a) \\\frac{(\sqrt{x^2+1}+x)^2+1}{2(\sqrt{x^2+1}+x)}=\\\frac{(sec(a)+tan(a))^2+1}{2(sec(a)+tan(a))}=\\\frac{(sec^2(a)+tan^2(a)+2sec(a)tan(a))+1}{2(sec(a)+tan(a)}=\\\frac{(sec^2(a)+tan^2(a)+2sec(a)tan(a))+1}{2(sec(a)+tan(a))}=\\as-we-know-(sec^2(a)=1+tan^2(a))\\so\\\frac{(sec^2(a)+sec^2(a)+2sec(a)tan(a))}{2(sec(a)+tan(a)}=\\\frac{2sec^2(a)+2sec(a)tan(a)}{2(sec(a)+tan(a)}=\\\frac{2sec^2(a)+2sec(a)tan(a))}{2(sec(a)+tan(a))}=\\\frac{2sec(a)(sec(a)+tan(a)}{2(sec(a)+tan(a))}=\\sec(a)=\sqrt{x^2+1}
$$
</s>