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# If $y=2\tan^{-1}x+\sin^{-1}\frac{2x}{1+x^2}$ for all x,then _________< y < ___________.

Toolbox:
• Put $x=tan\theta \rightarrow \theta = tan^{-1}x$
• $\large\frac{2tan\theta}{1+tan^2\theta}=sin2\theta$
• Principal value of tan is $(-\large\frac{\pi}{2},\large\frac{\pi}{2})$
Ans : $(-2\pi \: \: 2\pi)$

Put $x=tan\theta \Rightarrow \theta = tan^{-1}x$
$\Rightarrow\: \large\frac{2x}{1+x^2}=\large\frac{2tan\theta}{1+tan^2\theta}=sin2\theta$
$sin^{-1}\large\frac{2x}{1+x^2}=sin^{-1}sin2\theta=2\theta$

Substituting the values in the given exepression we get
$y=2.tan^{-1}tanx+sin^{-1}\large\frac{2x}{1+x^2}=2\theta+2\theta$

$\Rightarrow\: y = 4\theta=4\: tan^{-1}x$
$tan^{-1}x \: lies \: in \: (-\large\frac{\pi}{2} \: \large\frac{\pi}{2})$

$\Rightarrow y \in (-2\pi \: \: 2\pi)$

edited Mar 16, 2013