Question:
Simplify : \(\cot^{-1}\sqrt{\dfrac{1 + \cos 2x}{1 - \cos 2x}}\), \(x \in \left(0, \dfrac{\pi}{2}\right)\).
Simplify : \(\cot^{-1}\sqrt{\dfrac{1 + \cos 2x}{1 - \cos 2x}}\), \(x \in \left(0, \dfrac{\pi}{2}\right)\).
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Write 1 + cos 2x as 2cos^2 x and 1 - cos 2x as 2sin^2 x. The root becomes cot x, and cot^{-1}(cot x) = x on (0, pi/2).
Updated On: Oct 22, 2026
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Solution and Explanation
Step 1: Understanding the Concept:
We use the double-angle identities to turn the expression under the root into a perfect square. Then we use \(\cot^{-1}(\cot\theta) = \theta\) for \(\theta \in (0, \pi)\).
Step 2: Apply the identities.
We know \(1 + \cos 2x = 2\cos^2 x\) and \(1 - \cos 2x = 2\sin^2 x\).
\[ \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} = \sqrt{\frac{2\cos^2 x}{2\sin^2 x}} = \sqrt{\cot^2 x} = |\cot x| \]
Step 3: Remove the modulus.
For \(x \in \left(0, \frac{\pi}{2}\right)\), both \(\sin x\) and \(\cos x\) are positive. So \(\cot x > 0\) and \(|\cot x| = \cot x\).
Step 4: Use the principal value range.
The expression becomes \(\cot^{-1}(\cot x)\).
The principal range of \(\cot^{-1}\) is \((0, \pi)\). Our \(x\) lies in \(\left(0, \frac{\pi}{2}\right)\), which is inside \((0, \pi)\).
So \(\cot^{-1}(\cot x) = x\).
Final Answer:
The simplified form is \(x\). \[ \boxed{x} \]
We use the double-angle identities to turn the expression under the root into a perfect square. Then we use \(\cot^{-1}(\cot\theta) = \theta\) for \(\theta \in (0, \pi)\).
Step 2: Apply the identities.
We know \(1 + \cos 2x = 2\cos^2 x\) and \(1 - \cos 2x = 2\sin^2 x\).
\[ \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} = \sqrt{\frac{2\cos^2 x}{2\sin^2 x}} = \sqrt{\cot^2 x} = |\cot x| \]
Step 3: Remove the modulus.
For \(x \in \left(0, \frac{\pi}{2}\right)\), both \(\sin x\) and \(\cos x\) are positive. So \(\cot x > 0\) and \(|\cot x| = \cot x\).
Step 4: Use the principal value range.
The expression becomes \(\cot^{-1}(\cot x)\).
The principal range of \(\cot^{-1}\) is \((0, \pi)\). Our \(x\) lies in \(\left(0, \frac{\pi}{2}\right)\), which is inside \((0, \pi)\).
So \(\cot^{-1}(\cot x) = x\).
Final Answer:
The simplified form is \(x\). \[ \boxed{x} \]
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