Question:
If \(\cos^{-1} x - \sin^{-1} x = \frac{\pi}{6}\), then \(x\) is equal to
If \(\cos^{-1} x - \sin^{-1} x = \frac{\pi}{6}\), then \(x\) is equal to
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\(\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}\) for \(x \in [-1, 1]\).
Updated On: Apr 25, 2026
- \(\frac{\sqrt{3}}{2}\)
- \(\frac{1}{\sqrt{2}}\)
- \(\frac{\sqrt{3}}{4}\)
- \(-\frac{1}{\sqrt{2}}\)
- \(\frac{1}{2}\)
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The Correct Option is
Solution and Explanation
Step 1: Understanding the Concept:
Use the identity \(\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}\).
Step 2: Detailed Explanation:
Let \(\cos^{-1} x - \sin^{-1} x = \frac{\pi}{6}\)
Also \(\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}\)
Adding: \(2\cos^{-1} x = \frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3} \Rightarrow \cos^{-1} x = \frac{\pi}{3} \Rightarrow x = \cos\frac{\pi}{3} = \frac{1}{2}\)
Step 3: Final Answer:
\(x = \frac{1}{2}\).
Use the identity \(\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}\).
Step 2: Detailed Explanation:
Let \(\cos^{-1} x - \sin^{-1} x = \frac{\pi}{6}\)
Also \(\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}\)
Adding: \(2\cos^{-1} x = \frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3} \Rightarrow \cos^{-1} x = \frac{\pi}{3} \Rightarrow x = \cos\frac{\pi}{3} = \frac{1}{2}\)
Step 3: Final Answer:
\(x = \frac{1}{2}\).
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