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Mathematics Question on Inverse Trigonometric Functions

Let x∈(0,1).x \in \left(0,1\right). The set of all x such that sin−1x> cos−1x,sin^{-1} x >\, cos^{-1} x, is the interval:

A

(12,12)\left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)

B

(12,1)\left(\frac{1}{\sqrt{2}}, 1\right)

C

(0,1)

D

(0,32)\left(0,\frac{\sqrt{3}}{2}\right)

Answer

(12,1)\left(\frac{1}{\sqrt{2}}, 1\right)

Explanation

Solution

Given  sin−1x> cos−1x\,sin^{-1} x>\, cos^{-1} x where  x∈(0,1)\, x \in \left(0,1\right)
⇒sin−1x> π2−sin−1x\Rightarrow sin^{-1} x>\, \frac{\pi}{2}-sin^{-1} x
⇒2sin−1x> π2⇒sin−1x> π4\Rightarrow 2 sin^{-1} x>\, \frac{\pi}{2} \Rightarrow sin^{-1} x>\, \frac{\pi}{4}
⇒x> sinπ4⇒x> 12\Rightarrow x>\, sin \frac{\pi}{4} \Rightarrow x>\, \frac{1}{\sqrt{2}}
Maximum value of sin−1x is π2sin^{-1} x \, is \, \frac{\pi}{2}
So, maximum value of x is 1. So,
x∈(12,1).x \in \left(\frac{1}{\sqrt{2}} , 1\right).