<s>[INST] Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$ I've tried following this way, but I haven't succeeded. 
Thank you!
 [/INST] Calculate the derivative of both sides:
$$(\arctan x)'=\frac{1}{1+x^2}$$
$$\left(\arcsin\frac{x}{\sqrt{1+x^2}}\right)'=\frac{\sqrt{1+x^2}-\frac{x^2}{\sqrt{1+x^2}}}{1+x^2}\cdot\frac{1}{\sqrt{1-\frac{x^2}{1+x^2}}}=$$
$$=\frac{1}{(1+x^2)\sqrt{1+x^2}}\cdot\frac{\sqrt{1+x^2}}{\sqrt 1}=\frac{1}{1+x^2}$$
Since both derivatives are equal the functions are the same up to the sum of a constant:
$$\arctan x=\arcsin\frac{x}{\sqrt{1-x^2}}+C\,\,\,,\,\,C=\,\text{a constant}$$
Finally, to find what $\,C\,$ is you can, for example, input $\,x=0\,$ in the above...
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