Why is the argument of a hyperbolic function equal to twice the area of a hyperbolic sector? Most of us know that a unit circle is a circle with radius one and center at the origin.
Now let $O$ be the origin and $X$ be the $x$-intercept of the circle
Consider a point $R$ on that unit circle making an angle $t$ from the positive $X$ axis. If we calculate the area of the sector enclosed by the angle $t$ (the region $OXR$), that area would be $t/2$. But here is the big picture, that area we happened to calculate is actually half the angle we used to define the traditional trigonometric functions (sine, cosine, ...).
Now let us consider a unit hyperbola (assume that The orientation of the hyperbola is horizontal and here we only consider the right part of the hyperbola ignore the left side) centered at the origin.To make things clear let us label some points.
Let $O$ be the origin, $A$ be the $x$-intercept of the hyperbola, and $P$ be some point on the hyperbola.
We know that $PA$ is a curve. (When I say PA, I am not referring a line segment but part of our unit hyperbola). Now imagine the region $OPA$. The area of this region is half the argument we used to define hyperbolic functions (same way we used the area $OXR$ in the case of circle).
But wait. In the case of the circle above we showed (proof NOT shown HERE) that the area enclosed by the two radius and the arc of the circle is half of the angle $t$. But in the case of the hyperbola there is no such proof.

So how can we use something we are not sure of to our advantage? or are we just constructing hyperbolic functions in such a way that they take twice the area $OPA$ as an argument and give us the desired coordinates? or is it just to provide the same reasoning as the unit circle?

 A: You're basically asking "Why are hyperbolic radians defined as twice the area of a hyperbolic sector?"
Well, there is the nice conceptual connection with the circle. But, really, we don't have a choice. Consider ... 
Euler's formula tells us we can write $$\cos\theta = \frac12\left(e^{i\theta}+e^{-i\theta}\right) \qquad \sin\theta = \frac1{2i}\left(e^{i\theta}-e^{-i\theta}\right) \tag{1}$$ when $\theta$ is given in (circular) radians. In this instance, it doesn't matter whether "(circular) radians" are defined by exact length of a corresponding circular arc or by twice the area of the corresponding circular sector. Those radian calculations match, so take your pick.
Now it would be really convenient of the hyperbolic functions were given analogously by 
$$\cosh t = \frac12\left(e^{t}+e^{-t}\right) \qquad \sinh t = \frac1{2}\left(e^{t}-e^{t}\right) \tag{2}$$
This works precisely when $t$'s value in hyperbolic radians is taken to be twice the area of the corresponding hyperbolic sector. We don't have a choice in this case. And it's more-or-less a happy coincidence that the circular counterpart has a "twice sector area" interpretation for its radian measure; this makes for a satisfying unification that suggests that the twice-sector-area definition of radians is the natural definition for both cases.
