Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$ I've tried following this way, but I haven't succeeded. 
Thank you!

Let $\arctan x=y\Leftrightarrow x=\tan y$. Then,
$$\sin^2 y+\cos^2 y=1\Leftrightarrow \tan^2 y+1=\frac{1}{\cos^2 y}\Leftrightarrow \frac{1}{x^2+1}=1-\sin^2 y\Leftrightarrow \sin^2 y=\frac{x^2}{x^2+1}$$
and so $$\sin y= \frac{x}{\sqrt{1+x^2}}\Rightarrow \arctan x=y=\arcsin \frac{x}{\sqrt{1+x^2}}$$
