Let $f(x) = \sin^{-1}(\frac{2x}{1+x^2})$ Show that $f(x) = 2\tan^{-1}(x)$ 
Let $$f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~   -\infty<x<\infty.$$
Show that,
(a) $f(x) = 2\tan^{-1}(x)$ for $-1\leq x \leq 1$ and
(b) $f(x) = \pi-2\tan^{-1}(x)$ for $x \geq 1.$

Proof: I started off by equating $$\sin^{-1}\left(\frac{2x}{x^2+1}\right)=2\tan^{-1}(x)$$
(a)
We wish to show that these are equal for $-1\leq x \leq 1$.
For this domain $\displaystyle -1 \leq\frac{2x}{x^2+1} \leq 1 \implies -\frac{\pi}{2} \leq \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
$$\frac{2x}{x^2+1} = \sin(2\tan^{-1}(x))$$
The task is now to show $\sin(2\tan^{-1}(x))=\frac{2x}{x^2+1}$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
For $-1 \leq x \leq 1 \implies -\frac{\pi}{4}\leq\tan^{-1}(x)\leq\frac{\pi}{4} \implies -\frac{\sqrt{2}}{2}\leq \sin(\tan^{-1}(x)) \leq \frac{\sqrt{2}}{2}$
Also, $\frac{\sqrt{2}}{2}\leq\cos(\tan^{-1}(x)) \leq 1$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x)) \iff 2(\frac{x}{\sqrt{x^2+1}})(\frac{1}{\sqrt{x^2+1}})= \frac{2x}{x^2+1}$$
Which was to be shown.
(b)
$\displaystyle x \geq 1 \implies 0<\frac{2x}{1+x^2}\leq 1 \implies 0< \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
Hence,
We must show that $$\sin^{-1}(\frac{2x}{x^2+1}) = \pi - 2\tan^{-1}(x) \iff \frac{2x}{x^2+1} = \sin(\pi - 2\tan^{-1}(x))$$ for $x\geq 1$
$$\sin(\pi - 2\tan^{-1}(x))=\sin(\pi)\cos(2\tan^{-1}(x)) - \cos(\pi)\sin(2\tan^{-1}(x))=\sin(2\tan^{-1}(x))$$
$$\sin(2\tan^{-1}(x)) = 2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
Which was to be demonstrated.
Note: This problem didn't flow like I thought it would. I had imagined that during some of the intermediate steps, I would be presented with the option of choosing an $f(x)$ or trig function value that would only be true in one of the intervals. But no such situation presented itself. Did I do something wrong? Did I overlook something?
 A: $$
\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\theta\quad\iff\quad\sin(\theta)=\frac{2x}{1+x^2}\quad\land\quad-\frac\pi2\le\theta\le\frac\pi2
$$
Note that
$$
\cos(\theta)=\left|\frac{1-x^2}{1+x^2}\right|
$$

When $|x|\le1$,
$$
\tan(\theta/2)=\frac{\sin(\theta)}{1+\cos(\theta)}=\frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}}=x
$$
Therefore,
$$
\theta=2\tan^{-1}(x)
$$

When $|x|\gt1$,
$$
\tan(\theta/2)=\frac{\sin(\theta)}{1+\cos(\theta)}=\frac{\frac{2x}{1+x^2}}{1-\frac{1-x^2}{1+x^2}}=1/x
$$
Therefore,
$$
\theta=2\tan^{-1}(1/x)
$$

Thus,
$$
\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\left\{\begin{array}{}
2\tan^{-1}(x)&\text{if }|x|\le1\\
2\tan^{-1}(1/x)&\text{if }|x|\gt1
\end{array}\right.
$$
and
$$
\tan^{-1}(1/x)=\left\{\begin{array}{rl}
\frac\pi2-\tan^{-1}(x)&\text{if }x\gt0\\
-\frac\pi2-\tan^{-1}(x)&\text{if }x\lt0\\
\end{array}\right.
$$
A: As already commented by @DatBoi that the proof is incorrect, I would suggest you a different approach.

$ f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~   -\infty<x<\infty.$

a)  Substitute $x= \tan\theta$
$\implies f(x)= \sin^{-1}(\sin2\theta)$
$\implies f(x) =2\theta\quad   -\pi/2\leq2\theta\leq\pi/2$
$\implies f(x)= 2\tan^{-1}x \quad  -\pi/2\leq2\ (\tan^{-1}x)\leq\pi/2$
$\implies f(x) = 2\tan^{-1}x \quad  -\pi/4\leq\ (\tan^{-1}x)\leq\pi/4$
$\implies f(x)= 2\tan^{-1}x \quad -1\leq x\leq1$
Now, I hope you can do the second part yourself.
A: For part (a), I would suggest starting as $$\sin(y)=\frac{2x}{1+x}$$ Now, you can use this to build a right triangle with the side of $2x$ and the hypotenuse of $1+x$. Now you can find the other side (How?).
Now let's check the desired part: You need to calculate $tan(y/2)$ (you have $y=f(x)= 2 \tan^{-1} (x)$).
From half angle identities we know that $$\tan(y/2)=\frac{\sin(y/2)}{\cos(y/2)}$$ (Use half angle identities to get these $\sin(\frac{y}{2})$ and $\cos(\frac{y}{2})$ **).
Finally, you will end up $$\tan(\frac{y}{2})=x$$ which means $$f(x)=2 \tan^{-1} (x).$$
**You get those half angles by the right triangle that you got from $$\sin(y)=\frac{2x}{1+x}$$
A: Show that $f'(x)=\frac{2}{1+x^2}$, hence $f(x)$ and $2\arctan(x)$ differ only by a constant. Then substitute $x=0$ to show that this constant is $0$.
