Question

The value of $x$ which satisfies equation $2{\tan }^{-1}2x={\sin }^{-1}\frac{4x}{1+4x^2}$ is valid in the interval

• A. $[\frac{1}{2},\infty )$
• B. $(-\infty ,-\frac{1}{2}]$
• C. $[-1,1]$
• D. $[-\frac{1}{2},\frac{1}{2}]$

Answer 1

The correct option is D $[-\frac{1}{2},\frac{1}{2}]$

Given:

$2ta{n}^{-1}2x=si{n}^{-1}\frac{4x}{1+4x^2}$

$2x=tan\theta$

$x=\frac{tan\theta }{2}$

$2ta{n}^{-1}\left(tan\theta \right)=si{n}^{-1}\frac{2tan\theta }{1+ta{n}^2\theta }$

$2ta{n}^{-1}\left(tan\theta \right)=si{n}^{-1}\left(sin2\theta \right)$

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

$\frac{-\pi }{4}\le \theta \le \frac{\pi }{4}$

$-1\le tan\theta \le +1$

$\frac{-1}{2}\le \frac{tan\theta }{2}\le \frac{+1}{2}$

$\frac{-1}{2}\le x\le \frac{+1}{2}$