Use the formula [tex]1+tan^{2}y= \frac{1}{cos^{2}y} [/tex]. From the previous formula [tex]cos^{2}y= \frac{1}{1+tan^{2}y} [/tex] and since [tex]cos^2y+sin^2y=1[/tex] you'll have
[tex]sin^2y=1-cos^2y=1- \frac{1}{1+tan^2y} = \frac{tan^2y}{1+tan^2y} [/tex].
Since [tex]tan(arctan(x))=x[/tex], you obtain that [tex]sin^2y=sin^2(arctanx)= \frac{x^2}{1+x^2} [/tex] .
Answer: [tex]sin(arctanx)= \frac{x}{ \sqrt{1+x^2} } [/tex]
