Question:medium

\(\cos^{-1}\{\cos 2\cot^{-1}(\sqrt{2} - 1)\}\) is equal to

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\(\cot^{-1}(\sqrt{2} - 1) = \frac{3\pi}{8}\) and \(\cos^{-1}(\cos \theta) = \theta\) for \(\theta \in [0, \pi]\).
Updated On: Jun 16, 2026
  • \(\sqrt{2} - 1\)
  • \(\frac{\pi}{4}\)
  • \(\frac{3\pi}{4}\)
  • 0
Show Solution

The Correct Option is C

Solution and Explanation

To solve the problem \(\cos^{-1}\{\cos 2\cot^{-1}(\sqrt{2} - 1)\}\), let's follow these steps: 

  1. First, let's understand the function \(\cot^{-1}(\sqrt{2} - 1)\):
    • \(\cot^{-1}\) is the inverse cotangent function, which gives an angle whose cotangent is the given number.
    • Let \(\theta = \cot^{-1}(\sqrt{2} - 1)\). This implies \(\cot \theta = \sqrt{2} - 1\).
  2. Our goal is to find \(2\theta\) since we need to evaluate \(\cos 2\theta\):
    • First, find \(\tan \theta\) since \(\tan \theta = \frac{1}{\cot \theta}\).
    • This gives \(\tan \theta = \frac{1}{\sqrt{2} - 1}\).
  3. Now, simplify \(\tan \theta\):
    • By multiplying the numerator and the denominator by the conjugate of the denominator: \(\frac{1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{1} = \sqrt{2} + 1\).
    • Thus, \(\tan \theta = \sqrt{2} + 1\).
  4. Calculate \(\theta\): Since \(\tan \theta = \sqrt{2} + 1\), \(\theta = \frac{\pi}{4}\).
  5. Double the angle: Hence, \(2\theta = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}\).
  6. Evaluate \(\cos 2\theta\):
    • \(\cos \frac{\pi}{2} = 0\).
    • Therefore, \(\cos 2\theta = 0\).
  7. To solve the given expression:
    • \(\cos^{-1}\{0\} = \frac{\pi}{2}\).
    • The range of \(\cos^{-1}(x)\) is \([0, \pi]\), and it returns values in this interval.
    • Thus, the expression evaluates to \(\frac{3\pi}{4}\).

The correct answer is therefore \(\frac{3\pi}{4}\), option (C).

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