Question
Complete solution set of $${\tan ^2}\left( {{{\sin }^{ - 1}}x} \right) > 1$$ is
A.
$$\left( { - 1, - \frac{1}{{\sqrt 2 }}} \right) \cup \left( {\frac{1}{{\sqrt 2 }},1} \right)$$
B.
$$\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right) \sim \left\{ 0 \right\}$$
C.
$$\left( { - 1,1} \right) \sim \left\{ 0 \right\}$$
D.
None of these
Answer :
$$\left( { - 1, - \frac{1}{{\sqrt 2 }}} \right) \cup \left( {\frac{1}{{\sqrt 2 }},1} \right)$$
Solution :
$$\eqalign{
& {\tan ^2}\left( {{{\sin }^{ - 1}}x} \right) > 1 \cr
& \Rightarrow \frac{\pi }{4} < {\sin ^{ - 1}}x < \frac{\pi }{2}{\text{ or }} - \frac{\pi }{2} < {\sin ^{ - 1}}x < - \frac{\pi }{4} \cr
& \Rightarrow x \in \left( {\frac{1}{{\sqrt 2 }},1} \right){\text{ or }}x \in \left( { - 1, - \frac{1}{{\sqrt 2 }}} \right) \cr
& \Rightarrow x \in \left( { - 1, - \frac{1}{{\sqrt 2 }}} \right) \cup \left( {\frac{1}{{\sqrt 2 }},1} \right) \cr} $$