Question

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

• A. $\left\{1,2\right\}$
• B. $\left\{-1,2\right\}$
• C. $\left\{-1,1,0\right\}$
• D. $\left\{1,\frac{1}{2},0\right\}$

Answer 1

The correct option is C

$\left\{-1,1,0\right\}$

Explanation for the correct option:

Find the solution set of the given equation.

Since, $2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$.

So, ${\sin }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

We know that, ${\tan }^{-1}\left(x\right)={\sin }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$.

Therefore,

$$\begin{gathered} {\sin }^{-1}\left(x\right)={\sin }^{-1}\left(\frac{{\displaystyle \frac{2x}{1-x^2}}}{\sqrt{1+{\left({\displaystyle \frac{2x}{1-x^2}}\right)}^2}}\right) \\ \Rightarrow {\sin }^{-1}\left(x\right)={\sin }^{-1}\left(\frac{2x}{1+x^2}\right) \\ \Rightarrow x=\frac{2x}{1+x^2} \\ \Rightarrow x+x^3-2x=0 \\ \Rightarrow x^3-x=0 \\ \Rightarrow x\left(x^2-1\right)=0 \\ \Rightarrow x=-1,0,1 \end{gathered}$$

Therefore, the solution set of the given equation is $\left\{-1,1,0\right\}$.

Hence, option $C$ is the correct option.