How to "see" this algebraic transformation used to derive the exponential form of the hyperbolic cosine? In this page, the exponential form of cosh is derived from the geometric definition (which is based on the unit hyperbola). Here is one step of the derivation, fully expanded out for novices like me:
$$a = ln(b + \sqrt{b^2 - 1})$$
$$e^a = b + \sqrt{b^2 - 1}$$
$${1 \over e^a} = {1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = b + \sqrt{b^2 - 1} + {1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {b(b + \sqrt{b^2 - 1}) + \sqrt{b^2 - 1}(b + \sqrt{b^2 - 1}) + 1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {b^2 + b\sqrt{b^2 - 1} + b\sqrt{b^2 - 1} + b^2 - 1 + 1 \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = {2b^2 + 2b\sqrt{b^2 - 1} \over b + \sqrt{b^2 - 1}}$$
$$e^a + {1 \over e^a} = 2b$$
This is all well and good, but for the life of me I can't see: how would someone ever know in advance that adding $b + \sqrt{b^2 - 1}$ to its own reciprocal would clean things up so nicely? Is this just a "special case" which needs to be memorized, or is there some general principle behind it which can be used when solving similar equations?

Lets call $c=\sqrt{(b^2-1)}$. Note that $b^2-c^2=1$ so $(b+c)(b-c)=1$ and $b-c={1\over b+c}$ , so that $b+c +{1\over b+c} = b+c + b-c =2b$ 
