Question

The set of values of $x$, satisfying the equation ${\tan }^2\left({\sin }^{-1}x\right)>1$ is -

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

Answer 1

Answer: (C)

${\tan }^2\left({\sin }^{-1}x\right)>1$
either $\tan \left({\sin }^{-1}x\right)>1\Rightarrow {\sin }^{-1}x>{\tan }^{-1}1$
$\Rightarrow {\sin }^{-1}x>{\sin }^{-1}\frac{1}{\sqrt{2}}$
$\Rightarrow x>\frac{1}{\sqrt{2}}\Rightarrow \frac{1}{\sqrt{2}}<x<1$
or $\tan \left({\sin }^{-1}x\right)<-1\Rightarrow {\sin }^{-1}x<{\tan }^{-1}(-1)$
$\Rightarrow {\sin }^{-1}x<{\sin }^{-1}\left(-\frac{1}{\sqrt{2}}\right)\Rightarrow -1<x<-\frac{1}{\sqrt{2}}$
so $x\in (-1,1)-\left[\frac{-\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right]$