Square Roots of Hyperbolic Functions

Question

I have the following vector function:

$$ r(t)=(t-\sinh t\cosh t)\,\partial_x+2\cosh t\,\partial_y. $$

I computed its velocity to be as such:

$$ r'(t)=-2\sinh^2t\,\partial_x+2\sinh t\,\partial_y. $$

Therefore, its speed is as follows:

$$ v(t)=2\sinh t\cosh t. $$

However, when I computed this using Maple, the program (after simplifying) gave me this:

$$ v(t)=2\sqrt{\sinh^2t\cosh^2t}. $$

Why did it not get rid of the radical?

Edit: Here is the Maple code:

                                             

Answer 1

Probably because the speed is $$ 2 \; \cosh t \; \; | \sinh t | $$