$\pm$ sign in $y=\arcsin\frac{x}{\sqrt{1+x^2}}$ If: 
$$y=\arcsin\frac{x}{\sqrt{1+x^2}}$$
Then: 
$$\sin(y)=\frac{x}{\sqrt{1+x^2}}$$
$$\cos^2(y)=1-\sin^2(y)=\frac{1}{1+x^2}$$
$$ \tan^2(y)=\sec^2(y)-1=1+x^2-1=x^2$$
Therefore I would say:
$$\tan(y)=\pm x$$
However, my calculus book says (without the $\pm$):
$$\tan(y)=x$$
Question: Why can we remove the $\pm$?

The sign of $\tan(\theta)$ is not uniquely determined from an equation $\sin(\theta) = a$, which has two solutions with opposite signs for the tangent.   Under any convention for choosing one of the two $\theta$'s as the value of $\arcsin(a)$, the tangent is uniquely determined. The convention consistent with what you wrote is $\arcsin \in (-\pi/2, \pi/2]$
