Question

The solution set of the equation ${\sin }^{-1}x=2{\tan }^{-1}x$ is

• A. {1,2}
• B. {-1,2}
• C. {-1,1,0}
• D. {1,1/2,0}

Answer 1

Answer: (C)

We have,
${\sin }^{-1}x=2{\tan }^{-1}x$
${\sin }^{-1}x={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
$\left[\because 2{\tan }^{-1}A={\sin }^{-1}\left(\frac{2A}{1+A^2}\right)\right]$
$\Rightarrow x=\frac{2x}{1+x^2}$
$\Rightarrow x\left(1+x^2\right)=2x$
$\Rightarrow x+x^3-2x=0$
$\Rightarrow x^3-x=0$
$\Rightarrow x\left(x^2-1\right)=0$
$\Rightarrow x=0$ or $x^2-1=0$
$\Rightarrow x=0$ or $x^2-1=0$
$\Rightarrow x=0$ or $x^2=1$
$\Rightarrow x=0$ or $x=\pm 1$
$\therefore x\in \{0,-1,1\}$