If ${\cos ^{ - 1}}x > {\sin ^{ - 1}}x,\;{\rm{then}}$

$0 \le x < \frac{1}{{\sqrt 2 }}$

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

{\cos ^{ - 1}}x > {\sin ^{ - 1}}x,\;\mathrm{then}

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because, given the condition

{\cos ^{ - 1}}x > {\sin ^{ - 1}}x

, the solution lies in the interval

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

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