<s>[INST] Express $2\tan^{-1}x$ in terms of $\sin^{-1}$ and $\cos^{-1}$ $$
2\tan^{-1}x=\begin{cases}\tan^{-1}\frac{2x}{1-x^2} &\mbox{ if $|x|<1$}\\
\pi+\tan^{-1}\frac{2x}{1-x^2} &\mbox{ if $|x|>1$}
\end{cases}
$$
Similarly,

$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2} &\mbox{ if $|x|\leq1$}\\
\text{____________} &\mbox{ if $|x|>1$}
\end{cases}
$$

and

$$
2\tan^{-1}x=\begin{cases}\cos^{-1}\frac{1-x^2}{1+x^2} &\mbox{ if $x\geq0$}\\
\text{____________} &\mbox{ if $x<0$}
\end{cases}
$$

How do I derive the missing cases ?
My Attempt:
Take $\tan^{-1}x=A\implies x=\tan A$ where $\frac{-\pi}{2}<A<\frac{\pi}{2}\implies -\pi<2A<\pi$.
Case 1 :
$$
\cos2A=\frac{1-\tan^2A}{1+\tan^2A}=\frac{1-x^2}{1+x^2}=\cos\Big(\cos^{-1}\frac{1-x^2}{1+x^2}\Big)\\
\implies \cos^{-1}\frac{1-x^2}{1+x^2}=2n\pi\pm2A=2n\pi\pm2\tan^{-1}x
$$
As $0\leq\cos^{-1}\leq\pi$,
If $0\leq2A\leq\pi$
$$
\cos^{-1}\frac{1-x^2}{1+x^2}=2A=2\tan^{-1}x
$$
If $-\pi<2A<0$
$$
\cos^{-1}\frac{1-x^2}{1+x^2}= \color{red}{\text{?}}
$$
Case 2 :
$$
\sin2A=\frac{2\tan A}{1+\tan^2A}=\frac{2x}{1+x^2}=\sin\Big(\sin^{-1}\frac{2x}{1+x^2}\Big)\\
\implies \sin^{-1}\frac{2x}{1+x^2}=n\pi+(-1)^n.2A=n\pi+(-1)^n.2\tan^{-1}x
$$
As $\frac{-\pi}{2}\leq\sin^{-1}\leq\frac{\pi}{2}$,
If $\frac{\pi}{2}\leq2A\leq{\pi}$
$$
\sin^{-1}\frac{2x}{1+x^2}= \color{red}{\text{?}}
$$
what value should $\cos^{-1}\frac{1-x^2}{1+x^2}$ and $\sin^{-1}\frac{2x}{1+x^2}$ take and how do I proceed further ?
My Understanding:
For,
$$
\tan2A=\tan\big(\tan^{-1}\frac{2x}{1-x^2}\big)\implies\tan^{-1}\frac{2x}{1-x^2}=n\pi+2A
$$
If $\frac{\pi}{2}<2A<{\pi}$ I would take
$$
\tan^{-1}\frac{2x}{1-x^2}=-\pi+2A=-\pi+2\tan^{-1}x
$$
 [/INST] hint
If $A=\tan (\frac {x}{2}) $ then
$$\cos (x)=\frac {1-A^2}{1+A^2} $$
and
$$\sin (x)=\frac {2A}{1+A^2} $$
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