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Question

Mathematics Question on Inverse Trigonometric Functions

If (cos⁡−1x)2−(sin⁡−1x)2>0(\cos^{-1}x)^2-(\sin^{-1}x)^2>0, then

A

$x

B

$-1

C

$0\leq,x,

D

$-1\leq,x,

Answer

$-1\leq,x,

Explanation

Solution

(cos−1x)2−(sin−1x)2>0\left(cos^{-1} x\right)^{2}-\left(sin^{-1}x\right)^{2}>0 ⇒(cos−1x+sin−1x)(cos−1x−sin−1x)>0 \Rightarrow \left(cos^{-1}x+sin^{-1}x\right)\left(cos^{-1}x -sin^{-1}x\right) > 0 ⇒π2(π2−2 sin−1x)>0\Rightarrow \frac{\pi}{2}\left(\frac{\pi}{2} -2\,sin^{-1}x\right)>0 ⇒π2>2 sin−1x\Rightarrow \frac{\pi}{2}> 2 \,sin^{-1}x ⇒π4>sin−1x\Rightarrow \frac{\pi}{4}>sin^{-1} x i.e., sin−1x<π4sin^{-1}x < \frac{\pi}{4} Also −π2≤sin−1x≤π2-\frac{\pi}{2}\le sin^{-1} x \le\frac{\pi}{2} ∴−π2≤sin−1x<π4\therefore -\frac{\pi}{2} \le sin^{-1}x < \frac{\pi}{4} ⇒−1≤x<12 \Rightarrow -1 \le x < \frac{1}{\sqrt{2}}