Suppose [tex]y=\arctan x[/tex], where [tex]-\dfrac\pi2<y<\dfrac\pi2[/tex]. The restriction on the domain of [tex]y[/tex] is needed for the following step; over this interval, we can properly take the inverse of [tex]\arctan[/tex], i.e. take the tangent of both sides:
[tex]y=\arctan x\iff\tan y=\tan(\arctan x)=x[/tex]
Just to emphasize the point, we can only simplify [tex]\tan(\arctan x)[/tex] to [tex]x[/tex] as long as [tex]y=\arctan x[/tex] is smaller than [tex]\dfrac\pi2[/tex] in absolute value.
Now imagine you have a right triangle with a reference angle of [tex]y[/tex]. We're told that [tex]\tan y=x[/tex]. In this triangle, then, the lengths of its legs occur in a ratio of [tex]\dfrac x1[/tex]. In other words, if one of the legs had length [tex]1[/tex] (the leg adjacent to the angle), then the other must have length [tex]x[/tex] (the leg opposite the angle).
By the Pythagorean theorem, it would follow that the length of the hypotenuse is [tex]\sqrt{x^2+1^2}=\sqrt{x^2+1}[/tex]. Then the sine of this angle is
[tex]\sin(\arctan x)=\sin y=\dfrac x{\sqrt{x^2+1}}[/tex]
Personally, I wouldn't deem it necessary to rationalize the denominator, but we could go ahead and do it for giggles.
[tex]\dfrac x{\sqrt{x^2+1}}\cdot\dfrac{\sqrt{x^2+1}}{\sqrt{x^2+1}}=\dfrac{x\sqrt{x^2+1}}{x^2+1}[/tex]
