Question

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

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

Answer 1

The correct option is C $(-1,\frac{-1}{\sqrt{2}})\cup (\frac{1}{\sqrt{2}},1)$

$ta{n}^2\left(si{n}^{-1}\right(x\left)\right)>1$
Or
$tan\left(si{n}^{-1}\right(x\left)\right)<-1$ and $tan\left(si{n}^{-1}\right(x\left)\right)>1$
Consider
$tan\left(si{n}^{-1}\right(x\left)\right)<-1$
We get
$si{n}^{-1}\left(x\right)<ta{n}^{-1}(-1)$
Or
$si{n}^{-1}\left(x\right)<-\frac{\pi }{4}$
Or
$x<-\frac{1}{\sqrt{2}}$ ...(i)
Similarly
$tan\left(si{n}^{-1}\right(x\left)\right)>1$
Or
$si{n}^{-1}\left(x\right)>\frac{\pi }{4}$
Or
$x>\frac{1}{\sqrt{2}}$...(ii)
Also $sin\left(x\right)ϵ[-1,1]$...(iii)
Hence considering i,ii and iii, we get
$xϵ[-1,\frac{-1}{\sqrt{2}}]\cup [\frac{1}{\sqrt{2}},1]$