Complete solution set of t a n 2 ( s i n − 1 x ) > 1 is : Options: A . &nbsp ( − 1 − 1 √ 2 ) ∪ ( 1 √ 2 , 1 ) B . &nbsp ( − 1 √ 2 , 1 √ 2 ) − { 0 } C . &nbsp ( − 1 , 1 ) − { 0 } D . &nbsp None of these

Answer: Option A : A t a n 2 ( s i n − 1 x ) > 1 ⇒ t a n ( s i n − 1 x ) < − 1 or t a n ( s i n − 1 x ) > 1 ⇒ − π 2 < s i n − 1 x < − π 4 or π 4 < s i n − 1 x < π 2 ⇒ − 1 < x < − 1 √ 2 or 1 √ 2 < x < 1 ⇒ x ϵ ( − 1 − − 1 √ 2 ) ∪ ( 1 √ 2 , 1 )

Complete solution set of tan2(sin−1x)>1 is :

Discussion Forum : Inverse Trigonometric Functions

Question -

Complete solution set of $t a n^{2} (s i n^{- 1} x) > 1$ is :

Options:

A .

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

B .

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

C .

(- 1, 1) - {0}

D . None of these

Answer: Option A
:
A

t a n^{2} (s i n^{- 1} x) > 1

\Rightarrow t a n (s i n^{- 1} x) < - 1

t a n (s i n^{- 1} x) > 1

\Rightarrow - \frac{π}{2} < s i n^{- 1} x < - \frac{π}{4}

\frac{π}{4} < s i n^{- 1} x < \frac{π}{2}

\Rightarrow - 1 < x < \frac{- 1}{\sqrt{2}}

\frac{1}{\sqrt{2}} < x < 1

\Rightarrow x ϵ (- 1 - \frac{- 1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, 1)

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Discussion Forum : Inverse Trigonometric Functions

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