Question

If $y=2ta{n}^{-1}x+si{n}^{-1}\frac{2x}{1+x^2}$ then

• A. $-\pi /2<y<\pi /2$
• B. $-3\pi /2<y<3\pi /2$
• C. $-\pi <y<\pi$
• D. $-\pi /4<y<\pi /4$

Answer 1

Answer: (B)

$-3\pi /2<y<3\pi /2$

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

$-\frac{\pi }{2}<{\tan }^{-1}x<\frac{\pi }{2}\Rightarrow -\pi <2{\tan }^{-1}x<\pi$ ..(1)

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

Here $\theta =\frac{2x}{1+x^2}$

$-\frac{\pi }{2}\le {\sin }^{-1}\frac{2x}{1+x^2}\le \frac{\pi }{2}$ ..(2)

Adding (1) and (2), we get

$\Rightarrow -\frac{3\pi }{2}<2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}<\frac{3\pi }{2}$

$\Rightarrow -\frac{3\pi }{2}<y<\frac{3\pi }{2}$