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?

We can't bluntly take $\sin^{-1}(\sin 2y) = 2y$ and so with $\tan^{-1}(\tan 2y)$, because we don't know the value of $2y$ and the range in which it lies.

So, substitute directly.
$f(-10) = \sin^{-1}\left(\dfrac{-20}{101}\right)+\tan^{-1}\left(\dfrac{20}{99}\right) = -\sin^{-1}\left(\dfrac{20}{101}\right)+\tan^{-1}\left(\dfrac{20}{99}\right)$
Now let, $\tan z = \dfrac{20}{99} = \dfrac{20/101}{99/101}\Rightarrow \sin z =\dfrac{20}{101} \Rightarrow z = \sin^{-1}\left(\dfrac{20}{101}\right)$
(Here $0<\tan^{-1}\left(\dfrac{20}{99}\right)<\dfrac{\pi}2$)
So, we have $f(-10) = -\sin^{-1}\left(\dfrac{20}{101}\right)  + \sin^{-1}\left(\dfrac{20}{101}\right) = 0$
