<s>[INST] If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
 [/INST] From showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$
$$2\arctan x=\begin{cases}\arctan\dfrac{2x}{1-x^2}\text{ if } x^2\le1\\\pi+\arctan\dfrac{2x}{1-x^2}\text{ if } x^2>1\end{cases}$$
Again,  from  Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
$$\sin^{-1}(\sin2y)=\begin{cases}2y\text{ if } -\dfrac\pi2\le2y\le\dfrac\pi2\\\pi-2y\text{ if } 2y>\dfrac\pi2\iff \tan y>1\\ -\pi-2y\text{ if } 2y<-\dfrac\pi2\iff \tan y<-1\end{cases}$$
</s>