Question

If $y=2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}$ then

• A. $-\frac{\pi }{2}<y<\frac{\pi }{2}$
• B. $-3\frac{\pi }{2}<y<\frac{\pi }{2}$
• C. $-\pi <y<\pi$
• D. $-\frac{\pi }{4}<y<\frac{\pi }{4}$

Answer 1

Answer: (C)

$-\pi <y<\pi$

Let $x=\tan \theta$

Then we have,

$\Rightarrow y=2{\tan }^{-1}\left(\tan \theta \right)+{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)$

$\Rightarrow y=2{\tan }^{-1}\left(\tan \theta \right)+{\sin }^{-1}\left(\sin 2\theta \right)$

$\Rightarrow y=2\theta +2\theta$

$\Rightarrow y=4\theta$

But we know that $\frac{-\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$

$\Rightarrow \frac{-\pi }{2}\le {\sin }^{-1}\left(\sin 2\theta \right)\le \frac{\pi }{2}$

$\Rightarrow -1\le \sin 2\theta \le 1$

$\Rightarrow \frac{-\pi }{2}\le 2\theta \le \frac{\pi }{2}$

$\Rightarrow -\pi \le 4\theta \le \pi$

$\Rightarrow -\pi \le y\le \pi$