Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$

I’ve tried following this way, but I haven’t succeeded.

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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…

Consider the right angled triangle with sides $1,x,\sqrt{1+x^2}$

Let $\phi$ be the angle opposite to the side of length $x$.

We find that:
$$\phi=\arcsin(x/\sqrt{1+x^2})$$
$$\phi=\arctan(x/1)$$

Thus:
$$\arcsin(x/\sqrt{1+x^2})=\arctan(x)$$

As soon as you see $\arctan x$, draw a right triangle in which the “opposite” side has length $x$ and the “adjacent” side has length $1$. Then the angle to which those are “opposite” and “adjacent” is $\arctan x$.

The Pythagorean theorem then tells you the length of the hypotenuse.

That gives you the sine of the angle, since $\sin=\dfrac{\mathrm{opp}}{\mathrm{hyp}}$.

That tells you what the angle in question is the arcsine of.

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 $$\sin y= \frac{x}{\sqrt{1+x^2}}\Rightarrow \arctan x=y=\arcsin \frac{x}{\sqrt{1+x^2}}$$

put $$x=\tan(\theta)$$ Now rewrite the formula in $\theta$ instead of $x$. All you need, really, are these:
$$\tan(x)=\sin(x)/\cos(x)$$
$$\sin^2(x)+\cos^2(x)=1$$
should I be more explicit?

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, $\displaystyle\arctan x= y=\arcsin\frac x{\sqrt{x^2+1}}=\arccos\frac1{\sqrt{x^2+1}}$

To prove that
$$\sin(\arctan x)=\frac{x}{(1+x^2)^{\frac12}}$$
let $a=\arctan x$.
Then
$$s=\sin a=\tan a\cdot\cos a=\tan a(1-(\sin a)^2)^{\frac12}$$

$$s^2=(x^2)\cdot(1-s^2)$$

$$s^2=\frac{x^2}{1+x^2}$$

$$s=\sin a=\sin a(\arctan x)=\frac{x}{(1+x^2)^{\frac12}}$$