Question

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

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

Answer 1

Answer: (C)

$(-1,1)-[-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}]$

Given,

${\tan }^2\left({\sin }^{-1}x\right)>1$

either

$\tan \left({\sin }^{-1}x\right)>1\Rightarrow {\sin }^{-1}x>{\tan }^{-1}1$

$\Rightarrow {\sin }^{-1}x>{\sin }^{-1}\frac{1}{\sqrt{2}}$

$\Rightarrow x>\frac{1}{\sqrt{2}}\Rightarrow \frac{1}{\sqrt{2}}<x<1$ $\dots \left(1\right)$

$\tan \left({\sin }^{-1}x\right)<-1\Rightarrow {\sin }^{-1}x<{\tan }^{-1}(-1)$

$\Rightarrow {\sin }^{-1}x<{\sin }^{-1}[-\frac{1}{\sqrt{2}}]\Rightarrow -1<x<-\frac{1}{\sqrt{2}}$ $\dots \left(2\right)$

combining $\left(1\right)$ and $\left(2\right)$

so $x\in (-1,1)-[-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}]$