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

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

We have,

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

$⇒ tan^2 (sin^{-1}x)-1 > 0 $

$⇒tan(sin^{-1}x) < -1 $ or, $ tan(sin^{-1}x)>1$

$⇒ ∞ <tan (sin^{-1}x) < -1 $ or $1 < tan (sin^{-1}x) < ∞$

$⇒ -\frac{\pi}{2} < (sin^{-1}x) < -\frac{\pi}{4}$ or, $ \frac{\pi}{4} < sin^{-1} x < \frac{\pi}{2}$

$⇒ x ∈ \left(-1, -\frac{1}{\sqrt{2}}\right)$ or, $x ∈ \left(\frac{1}{\sqrt{2}}, 1\right)$

$⇒ x ∈ (-1, 1) - \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$