If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$

Question

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?

Answer 1

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 $f(-10) = -\sin^{-1}\left(\dfrac{20}{101}\right) + \sin^{-1}\left(\dfrac{20}{101}\right) = 0$

Answer 2

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 $1-x^2<0\iff x^2>1, u=\sin^{-1}\left(-\dfrac{2x}{1+x^2}\right)=-\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$

Answer 3

$$\arcsin(\sin(x))=\left\{ \begin{array}{ll} x-2\pi n&\hbox{for }-\frac{\pi}{2}+2\pi n\le x\le \frac{\pi}{2}+2\pi n\\ \pi-x-2\pi n&\hbox{for }\frac{\pi}{2}+2\pi n\le x\le \frac{3\pi}{2}+2\pi n\\ \end{array}\right.,$$ $$\arctan(\tan(x))=x-\left\lfloor\frac{x+\pi/2}{\pi}\right\rfloor\pi$$

Answer 4

Hint:

$$\sin^{-1}\dfrac{2(-10)}{1+(-10)^2}=\sin^{-1}\left(-\dfrac{20}{101}\right)$$

$$\tan^{-1}\dfrac{2(-10)}{1-(-10)^2}=\tan^{-1}\dfrac{20}{99}=u(\text{say})$$

$\implies\dfrac\pi2>u>0$

$\sec u=+\sqrt{1+\left(\dfrac{20}{99}\right)^2}=\dfrac{101}{99}$

$\implies\sin u=\dfrac{\tan u}{\sec u}=?$

$u=\sin^{-1}?$

Answer 5

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 $

$$\sin^{-1}(\sin2y)=\begin{cases}2y\text{ if } -\dfrac\pi2\le2y\le\dfrac\pi2\\\pi-2y\text{ if } 2y>\dfrac\pi2\iff \tan y>1\\ -\pi-2y\text{ if } 2y<-\dfrac\pi2\iff \tan y<-1\end{cases}$$