In proving $\cosh(\sinh^{-1}(x))=\sqrt{1+x^2}$ don't we need restrictions on values of $x$? 


*The functions $\sinh$ and $\tanh$ are one-one; their inverses $\sinh^{-1}$ and $\tanh^{-1}$ are defined on $\mathbb{R}$ and $(-1,1)$, respectively. These inverse functions are sometimes denoted by $\arg\sinh$ and $\arg\tanh$ (the "argument" of the hyperbolic sine and tangent.
If $\cosh$ is restricted to $[0,\infty)$ it has an inverse, denoted by
$\arg\cosh$, or simply $\cosh^{-1}$, which is defined on $[1,\infty)$.

Prove, using information from problem $8$, that
(b) $\cosh(\sinh^{-1}(x))=\sqrt{1+x^2}\tag{1}$

Shouldn't there be a restriction on the values that $x$ can take in $(1)$ (such restrictions are present in other items in this problem, for example)?
In problem $8$ we proved that
$$\cosh^2-\sinh^2=1\tag{1}$$
Hence, expression $(1)$ evaluated at a point $\sinh^{-1}(x)$ is
$$\cosh^2(\sinh^{-1}(x))-\sinh^2(\sinh^{-1}(x))=1$$
$$\cosh^2(\sinh^{-1}(x))-x^2=1$$
$$\cosh^2(\sinh^{-1}(x))=1+x^2\tag{2}$$
$$\cosh(\sinh^{-1}(x))=\sqrt{1+x^2}\tag{3}$$
Consider the step from $(2)$ to $(3)$.
In $(2)$ don't we need to restrict $\sinh^{-1}(x)$ to be in $[1,\infty)$?
Ie
$$\sinh^{-1}(x)\in [1,\infty)$$
Now, since $\sinh$ is increasing and
$$\sinh^{-1}(x)=1 \implies x=\sinh{1}$$
then if $x\in[\sinh{1},\infty)$ then $\sinh^{-1}(x) \in [1,\infty)$.
Finally, just to confirm, the reason we take the positive square root is because $\cosh$ always positive, correct?
 A: The domain of $\cosh^{-1}$ may be $[1,\infty)$, but there is no $\cosh^{-1}$ in the identity you're talking about so this interval is irrelevant. The domain of $\sinh^{-1} x$ is all real numbers; its range is all real numbers; the domain of $\cosh$ is all real numbers; so $\cosh(\sinh^{-1}x)$ is well-defined for all $x$.

Shouldn't there be a restriction on the values that $x$ can take in $\cosh(\sinh^{-1}(x))=\sqrt{1+x^2}$ (such restrictions are present in other items in this problem, for example)?

No.

In $\cosh^2(\sinh^{-1}(x))=1+x^2$ don't we need to restrict $\sinh^{−1}(x)$ to be in $[1,\infty)$?

No.

Finally, just to confirm, the reason we take the positive square root is because cosh always positive, correct?

Yes.
A: The formula holds without any restrictions, since $sinh(x)$ is strictly increasing and hence invertible on $\mathbb{R}$. Let $sinh^{-1}(x)=y$. Then $x=sinhy=\dfrac{e^{y}-e^{-y}}{2}$ .
Soving the binomial for $e^{y}$ we get $e^{y}=x\pm \sqrt{x^{2}+1}$. Since $e^{y}$ is positive, the minus case is rejected and we obtain $e^{y}=x+\sqrt{x^{2}+1}$ and $y=ln(x+\sqrt{x^{2}+1})$.
Now $cosh(y)$=$\dfrac{x+\sqrt{x^{2}+1}}{2}$ +$\dfrac{1}{2(x+\sqrt{x^{2}+1})}$=$\sqrt{x^{2}+1}$.
Thus no restrictions on $x$ are required!!
