Question

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

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

Answer 1

Answer: (C)

We have,
$\Rightarrow {\tan }^2\left({\sin }^{-1}x\right)>1$
$\Rightarrow {\tan }^2\left({\sin }^{-1}x\right)-1>0$
$\Rightarrow \tan \left({\sin }^{-1}x\right)<-1$
or, $\tan \left({\sin }^{-1}x\right)>1$
$\Rightarrow -\infty <\tan \left({\sin }^{-1}x\right)<-1$
or, $1<\tan \left({\sin }^{-1}x\right)<\infty$
$\Rightarrow -\frac{\pi }{2}<{\sin }^{-1}x<-\frac{\pi }{4}$
or, $\frac{\pi }{4}<{\sin }^{-1}x<\frac{\pi }{2}$
$\Rightarrow x\in \left(-1,-\frac{1}{\sqrt{2}}\right)$
or, $x\in \left(\frac{1}{\sqrt{2}},1\right)$
$\Rightarrow x\in (-1,-1)-\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$