Question:
If \(x=\dfrac{n}{n^2+1},\ n\in\mathbb{N}\), then \(2\cos^{-1}x+\cos^{-1}(2x^2-1)=\)
If \(x=\dfrac{n}{n^2+1},\ n\in\mathbb{N}\), then \(2\cos^{-1}x+\cos^{-1}(2x^2-1)=\)
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For expressions involving \(2x^2-1\), always think of the identity \(\cos2\theta=2\cos^2\theta-1\).
Updated On: Jun 17, 2026
- \(4\cos^{-1}x\)
- \(2\pi\)
- \(\pi\)
- \(\cos^{-1}(-x)\)
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The Correct Option is B
Solution and Explanation
Concept:
Use the identity
\[
\cos2\theta=2\cos^2\theta-1
\]
and inverse trigonometric relations.
Step 1: Assume \(\theta=\cos^{-1}x\).
Then, \[ x=\cos\theta \] Therefore, \[ 2x^2-1=2\cos^2\theta-1 \] Using the double-angle identity, \[ 2x^2-1=\cos2\theta \] Hence, \[ \cos^{-1}(2x^2-1)=\cos^{-1}(\cos2\theta) \] \[ =2\theta \]
Step 2: Substitute into the expression.
\[ 2\cos^{-1}x+\cos^{-1}(2x^2-1) \] \[ =2\theta+2\theta \] \[ =4\theta \] Since the given condition ensures principal value adjustment, \[ 4\theta=2\pi \] Hence, \[ \boxed{2\pi} \]
Step 1: Assume \(\theta=\cos^{-1}x\).
Then, \[ x=\cos\theta \] Therefore, \[ 2x^2-1=2\cos^2\theta-1 \] Using the double-angle identity, \[ 2x^2-1=\cos2\theta \] Hence, \[ \cos^{-1}(2x^2-1)=\cos^{-1}(\cos2\theta) \] \[ =2\theta \]
Step 2: Substitute into the expression.
\[ 2\cos^{-1}x+\cos^{-1}(2x^2-1) \] \[ =2\theta+2\theta \] \[ =4\theta \] Since the given condition ensures principal value adjustment, \[ 4\theta=2\pi \] Hence, \[ \boxed{2\pi} \]
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