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Q.

Complete solution set of $\tan^{2} (\sin^{- 1} x) > 1$ is

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a

$(- 1, 1) ~ {0}$

b

none of these

c

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

d

$(- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) - {0}$

answer is A.

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Detailed Solution

${[\tan (\sin^{- 1} x)]}^{2} > 1$
$\Rightarrow \tan (\sin^{- 1} x) < - 1 or \tan (\sin^{- 1} x) > 1$
$\begin{array}{l} \Rightarrow \frac{π}{4} < \sin^{- 1} x < \frac{π}{2} or - \frac{π}{2} < \sin^{- 1} x < - \frac{π}{4} \\ \Rightarrow x \in (- 1, - \frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, 1) \end{array}$

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