$\pm$ 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 $\pm$?

We can get rid of the $\pm$ sign because in $y=\arcsin\frac{x}{\sqrt{1+x^2}}$, $x$ and $y$ have to have the same sign: For $-\pi/2 < y\le\pi/2$ if $x$ is positive, then $y$ is positive then also $\tan(y)$ is positive. If $x$ is negative then $y$ is negative then $\tan(y)$ is negative.
Therefore, if $ \tan^2(y)=x^2$ then $ \tan(y)=x$
