<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] Hint:
$$\sin^{-1}\dfrac{2(-10)}{1+(-10)^2}=\sin^{-1}\left(-\dfrac{20}{101}\right)$$
$$\tan^{-1}\dfrac{2(-10)}{1-(-10)^2}=\tan^{-1}\dfrac{20}{99}=u(\text{say})$$
$\implies\dfrac\pi2>u>0$
$\sec u=+\sqrt{1+\left(\dfrac{20}{99}\right)^2}=\dfrac{101}{99}$
$\implies\sin u=\dfrac{\tan u}{\sec u}=?$
$u=\sin^{-1}?$
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