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: \[ x = \frac{1}{\sqrt{6}}. \]