If $y=2 an^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2} ight)$ for all $x$, then ____ $

Question

If $y=2	an^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}ight)$ for all $x$, then ____ $<y<$ ____.

Accepted Answer

> [!TLDR] > Case-splitting the identity for $\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)$ gives $y=4\tan^{-1}x$ on $|x|\le1$, $y=\pi$ for $x>1$, and $y=-\pi$ for $x<-1$; so $y$ ranges over $[-\pi,\pi]$ and the blanks are $-\pi$ and $\pi$. ## The idea $\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)$ equals $2\tan^{-1}x$ only while $|x|\le1$; beyond that the output is folded back into $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, so the identity picks up a $\pm\pi$. We handle the three regions separately. ## Step 1 — The identity, by cases With $x=\tan\theta$, $\frac{2x}{1+x^2}=\sin2\theta$, and reducing $2\theta$ into $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ gives $$\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\begin{cases}2\tan^{-1}x,&|x|\le1\[2pt]\pi-2\tan^{-1}x,&x>1\[2pt]-\pi-2\tan^{-1}x,&x<-1.\end{cases}$$ ## Step 2 — Form $y$ in each region **$|x|\le1$:** $y=2\tan^{-1}x+2\tan^{-1}x=4\tan^{-1}x$. As $x$ increases from $-1$ to $1$, $\tan^{-1}x$ increases from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$, so $y$ increases continuously from $-\pi$ to $\pi$. **$x>1$:** $y=2\tan^{-1}x+\pi-2\tan^{-1}x=\pi$ (constant). …

Q.If $y=2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ for all $x$ , then $<y<$ .

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

The idea

Step 1 — The identity, by cases

Step 2 — Form $y$ in each region

Method: Case-split the identity for $\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)$

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.If y=2tan⁡−1x+sin⁡−1(2x1+x2)y=2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)y=2tan−1x+sin−1(1+x22x​) for all xxx, then ____ <y<<y<<y< ____.

The idea

Step 1 — The identity, by cases

Step 2 — Form yyy in each region

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Q.If $y=2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ for all $x$ , then $<y<$ .

Step 2 — Form $y$ in each region

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