Question

The value of \( \sin(\cot^{-1}(x)) \) is:

• A. \( \frac{1}{\sqrt{1 + x^2}} \)
• B. \( \sqrt{1 + x^2} \)
• C. \( \frac{1}{x\sqrt{1 + x^2}} \)
• D. \( x\sqrt{1 + x^2} \)

Hint:

For trigonometric inverses, visualize right triangles to determine relationships.

Answer 1

The Correct Option is A

\begin{enumerate} \item Let \( \cot^{-1}(x) = \theta \). By the definition of inverse trigonometric functions, \( \cot(\theta) = x \). \item Recall the trigonometric identity for cotangent: \[ \cot(\theta) = \frac{\text{adjacent}}{\text{opposite}}. \] Therefore, if \( \cot(\theta) = x \), we can write: \[ \text{adjacent} = x, \quad \text{opposite} = 1. \] \item Use the Pythagorean theorem to find the hypotenuse of the right triangle: \[ \text{hypotenuse} = \sqrt{\text{adjacent}^2 + \text{opposite}^2} = \sqrt{x^2 + 1}. \] \item Now, calculate \( \sin(\theta) \). By definition: \[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}. \] Substituting the values: \[ \sin(\theta) = \frac{1}{\sqrt{x^2 + 1}}. \] \item Hence, the value of \( \sin(\cot^{-1}(x)) \) is: \[ \boxed{\frac{1}{\sqrt{1 + x^2}}}. \] \end{enumerate}