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

Let $\displaystyle\arctan x= y$
$\implies(i) \tan y =x$ 
and $(ii)\displaystyle-\frac\pi2\le y\le\frac\pi2$ (using the definition of principal value)
$\implies \cos y\ge0$
We have $$\frac{\sin y}x=\frac {\cos y }1=\pm\sqrt{\frac{\sin^2y+\cos^2y}{x^2+1^2}}=\pm\frac1{\sqrt{x^2+1}}$$
$\displaystyle\implies \cos y=+\frac1{\sqrt{x^2+1}}$ and $\displaystyle\sin y=\frac x{\sqrt{x^2+1}}$
So, $\displaystyle\arctan x= y=\arcsin\frac x{\sqrt{x^2+1}}=\arccos\frac1{\sqrt{x^2+1}}$
