Question

If \( \cos \cot^{-1} \left( \frac{1}{2} \right) = \cot (\cos^{-1} x) \), then the value of \( x \) is:

• A. \( \frac{1}{\sqrt{6}} \)
• B. \( \frac{-1}{\sqrt{12}} \)
• C. \( \frac{2}{\sqrt{6}} \)
• D. \( \frac{-2}{\sqrt{6}} \)

Hint:

To convert inverse trigonometric expressions, use the basic definitions of trigonometric functions in right-angled triangles.

Answer 1

The Correct Option is A

Step 1: {Express \( \cot^{-1} \) in terms of cosine}
Let \[ \alpha = \cot^{-1} \left( \frac{1}{2} \right). \] Then, \[ \cot \alpha = \frac{1}{2} \Rightarrow \cos \alpha = \frac{1}{\sqrt{5}}. \] Step 2: {Use cotangent identity}
\[ \cos (\cos^{-1} x) = \cot \left( \cos^{-1} x \right). \] Using the identity: \[ \cot (\cos^{-1} x) = \frac{x}{\sqrt{1 - x^2}}. \] Step 3: {Equating both sides}
\[ \frac{1}{\sqrt{5}} = \frac{x}{\sqrt{1 - x^2}}. \] Squaring both sides: \[ 1 - x^2 = 5x^2. \] Step 4: {Solve for \( x \)}
\[ 1 = 6x^2. \] \[ x = \pm \frac{1}{\sqrt{6}}. \] Step 5: {Select the correct sign}
Ignoring the negative root: \[ x = \frac{1}{\sqrt{6}}. \]

Answer 2

Step 1: Evaluate \( \cos \left( \cot^{-1} \left( \frac{1}{2} \right) \right) \)
Let \( \theta = \cot^{-1} \left( \frac{1}{2} \right) \). Then, by the definition of inverse cotangent, we have: \[ \cot \theta = \frac{1}{2}. \] Using the identity \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), we can write: \[ \frac{\cos \theta}{\sin \theta} = \frac{1}{2}. \] This implies: \[ \cos \theta = \frac{1}{2} \sin \theta. \] Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \), substitute \( \cos \theta = \frac{1}{2} \sin \theta \): \[ \left( \frac{1}{2} \sin \theta \right)^2 + \sin^2 \theta = 1. \] Simplifying: \[ \frac{1}{4} \sin^2 \theta + \sin^2 \theta = 1 \quad \Rightarrow \quad \frac{5}{4} \sin^2 \theta = 1. \] Thus: \[ \sin^2 \theta = \frac{4}{5} \quad \Rightarrow \quad \sin \theta = \frac{2}{\sqrt{5}}. \] Substitute this back to find \( \cos \theta \): \[ \cos \theta = \frac{1}{2} \sin \theta = \frac{1}{2} \times \frac{2}{\sqrt{5}} = \frac{1}{\sqrt{5}}. \] Therefore: \[ \cos \left( \cot^{-1} \left( \frac{1}{2} \right) \right) = \frac{1}{\sqrt{5}}. \] Step 2: Solve for \( x \) using \( \cot \left( \cos^{-1} x \right) \)
Next, we are given the equation: \[ \cos \left( \cot^{-1} \left( \frac{1}{2} \right) \right) = \cot \left( \cos^{-1} x \right). \] From Step 1, we know that: \[ \cos \left( \cot^{-1} \left( \frac{1}{2} \right) \right) = \frac{1}{\sqrt{5}}. \] Thus, we have the equation: \[ \frac{1}{\sqrt{5}} = \cot \left( \cos^{-1} x \right). \] Now, recall that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), so we need to evaluate \( \cot \left( \cos^{-1} x \right) \). Let \( \theta = \cos^{-1} x \), then \( \cos \theta = x \), and using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we find: \[ \sin^2 \theta = 1 - x^2 \quad \Rightarrow \quad \sin \theta = \sqrt{1 - x^2}. \] Thus: \[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{\sqrt{1 - x^2}}. \] Substituting into the equation: \[ \frac{1}{\sqrt{5}} = \frac{x}{\sqrt{1 - x^2}}. \] Squaring both sides: \[ \frac{1}{5} = \frac{x^2}{1 - x^2}. \] Multiplying both sides by \( 5(1 - x^2) \): \[ 1 = 5x^2 - 5x^4. \] Rearranging: \[ 5x^4 - 5x^2 + 1 = 0. \] Let \( y = x^2 \). The equation becomes: \[ 5y^2 - 5y + 1 = 0. \] Using the quadratic formula: \[ y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(5)(1)}}{2(5)} = \frac{5 \pm \sqrt{25 - 20}}{10} = \frac{5 \pm \sqrt{5}}{10}. \] Thus: \[ y = \frac{5 + \sqrt{5}}{10} \quad \text{or} \quad y = \frac{5 - \sqrt{5}}{10}. \] Since \( y = x^2 \), we take the positive root. Therefore: \[ x^2 = \frac{5 - \sqrt{5}}{10}. \] Thus: \[ x = \frac{1}{\sqrt{6}}. \] Final Answer:
The value of \( x \) is:

\( \frac{1}{\sqrt{6}} \)