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?
 A: 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 $f(-10) = -\sin^{-1}\left(\dfrac{20}{101}\right)  + \sin^{-1}\left(\dfrac{20}{101}\right) = 0$
A: Let $\tan^{-1}\dfrac{2x}{1-x^2}=u\implies-\dfrac\pi2<u<\dfrac\pi2$
$\tan u=\dfrac{2x}{1-x^2}$
$\implies\sec u+\sqrt{1+\left(\dfrac{2x}{1-x^2}\right)^2}=\dfrac{1+x^2}{|1-x^2|}$
$\sin u=\dfrac{\tan u}{\sec u}=\text{sign of}(1-x^2)\cdot\dfrac{2x}{1+x^2}$
$\implies u=\sin^{-1}\left(\text{sign of}(1-x^2)\cdot\dfrac{2x}{1+x^2}\right)$
So if $1-x^2<0\iff x^2>1, u=\sin^{-1}\left(-\dfrac{2x}{1+x^2}\right)=-\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$
A: $$\arcsin(\sin(x))=\left\{
\begin{array}{ll}
x-2\pi n&\hbox{for }-\frac{\pi}{2}+2\pi n\le x\le \frac{\pi}{2}+2\pi n\\
\pi-x-2\pi n&\hbox{for }\frac{\pi}{2}+2\pi n\le x\le \frac{3\pi}{2}+2\pi n\\
\end{array}\right.,$$
$$\arctan(\tan(x))=x-\left\lfloor\frac{x+\pi/2}{\pi}\right\rfloor\pi$$
A: Hint:
$$\sin^{-1}\dfrac{2(-10)}{1+(-10)^2}=\sin^{-1}\left(-\dfrac{20}{101}\right)$$
$$\tan^{-1}\dfrac{2(-10)}{1-(-10)^2}=\tan^{-1}\dfrac{20}{99}=u(\text{say})$$
$\implies\dfrac\pi2>u>0$
$\sec u=+\sqrt{1+\left(\dfrac{20}{99}\right)^2}=\dfrac{101}{99}$
$\implies\sin u=\dfrac{\tan u}{\sec u}=?$
$u=\sin^{-1}?$
A: 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 $
$$\sin^{-1}(\sin2y)=\begin{cases}2y\text{ if } -\dfrac\pi2\le2y\le\dfrac\pi2\\\pi-2y\text{ if } 2y>\dfrac\pi2\iff \tan y>1\\ -\pi-2y\text{ if } 2y<-\dfrac\pi2\iff \tan y<-1\end{cases}$$
