Question:
If $\sin^{-1}\left(\frac{x}{1+x}\right) = \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{2}\right)$, then $x$ is equal to:
If $\sin^{-1}\left(\frac{x}{1+x}\right) = \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{2}\right)$, then $x$ is equal to:
Show Hint
Recognizing the $\sin^{-1} \theta + \cos^{-1} \theta = \pi/2$ identity saves having to calculate the actual angles.
Updated On: Apr 28, 2026
- $\frac{1}{2}$
- 2
- 3
- 1
- $\frac{1}{4}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Use the identity $\sin^{-1} \theta + \cos^{-1} \theta = \frac{\pi}{2} \implies \frac{\pi}{2} - \cos^{-1} \theta = \sin^{-1} \theta$.
Step 2: Analysis
$\sin^{-1}(\frac{x}{1+x}) = \sin^{-1}(\frac{1}{2})$. Therefore, $\frac{x}{1+x} = \frac{1}{2}$.
Step 3: Calculation
$2x = 1 + x \implies x = 1$. Final Answer: (D)
Use the identity $\sin^{-1} \theta + \cos^{-1} \theta = \frac{\pi}{2} \implies \frac{\pi}{2} - \cos^{-1} \theta = \sin^{-1} \theta$.
Step 2: Analysis
$\sin^{-1}(\frac{x}{1+x}) = \sin^{-1}(\frac{1}{2})$. Therefore, $\frac{x}{1+x} = \frac{1}{2}$.
Step 3: Calculation
$2x = 1 + x \implies x = 1$. Final Answer: (D)
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