Question:
If \( (\cos^{-1} x)^2 - (\sin^{-1} x)^2>0 \), then}
If \( (\cos^{-1} x)^2 - (\sin^{-1} x)^2>0 \), then}
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Use the identity $\sin^{-1} x + \cos^{-1} x = \pi/2$ to convert quadratic inequalities into linear ones.
Updated On: Apr 30, 2026
- \( x<\frac{1}{2} \)
- \( x>0 \)
- \( 0 \le x<\frac{1}{\sqrt{2}} \)
- \( -1 \le x<\frac{1}{\sqrt{2}} \)
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The Correct Option is D
Solution and Explanation
Step 1: Factorize
$(\cos^{-1} x - \sin^{-1} x)(\cos^{-1} x + \sin^{-1} x)>0$.
Step 2: Use Identity
$\cos^{-1} x + \sin^{-1} x = \pi/2$.
$(\cos^{-1} x - \sin^{-1} x)(\pi/2)>0 \implies \cos^{-1} x>\sin^{-1} x$.
Step 3: Solve Inequality
We know $\cos^{-1} x = \sin^{-1} x$ at $x = 1/\sqrt{2}$.
Since $\cos^{-1} x$ is decreasing and $\sin^{-1} x$ is increasing, $\cos^{-1} x>\sin^{-1} x$ when $x<1/\sqrt{2}$.
Step 4: Combine with Domain
Domain of inverse functions is $[-1, 1]$. So, $-1 \le x<1/\sqrt{2}$.
Final Answer:(D)
$(\cos^{-1} x - \sin^{-1} x)(\cos^{-1} x + \sin^{-1} x)>0$.
Step 2: Use Identity
$\cos^{-1} x + \sin^{-1} x = \pi/2$.
$(\cos^{-1} x - \sin^{-1} x)(\pi/2)>0 \implies \cos^{-1} x>\sin^{-1} x$.
Step 3: Solve Inequality
We know $\cos^{-1} x = \sin^{-1} x$ at $x = 1/\sqrt{2}$.
Since $\cos^{-1} x$ is decreasing and $\sin^{-1} x$ is increasing, $\cos^{-1} x>\sin^{-1} x$ when $x<1/\sqrt{2}$.
Step 4: Combine with Domain
Domain of inverse functions is $[-1, 1]$. So, $-1 \le x<1/\sqrt{2}$.
Final Answer:(D)
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