Question

Write \( \cot^{-1}\left\{\frac{1}{\sqrt{x^2-1}}\right\}; x>1 \) in the simplest form.

Hint:

Memorize the standard substitutions for simplifying inverse trigonometric functions:
For \( \sqrt{a^2 - x^2} \), use \( x = a\sin\theta \) or \( x = a\cos\theta \).
For \( a^2 + x^2 \) or \( \sqrt{a^2 + x^2} \), use \( x = a\tan\theta \) or \( x = a\cot\theta \).
For \( \sqrt{x^2 - a^2} \), use \( x = a\sec\theta \) or \( x = a\csc\theta \).

Answer 1

Step 1: Understanding the Concept:
The goal is to simplify the given inverse trigonometric expression. This is typically achieved by using a trigonometric substitution that simplifies the term inside the function, allowing us to use the property \( \cot^{-1}(\cot\theta) = \theta \). The presence of the term \( \sqrt{x^2 - 1} \) suggests the substitution \( x = \sec\theta \).
Step 2: Key Formula or Approach:
We will use the substitution \( x = \sec\theta \). This is chosen because of the trigonometric identity \( \sec^2\theta - 1 = \tan^2\theta \). The given condition \( x>1 \) implies that \( \sec\theta>1 \), which means \( \theta \) lies in the first quadrant, i.e., \( 0<\theta<\frac{\pi}{2} \). This is important for determining the sign when taking the square root.
Step 3: Detailed Calculation:
Let the given expression be \( y \). \[ y = \cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right) \] Substitute \( x = \sec\theta \). This implies \( \theta = \sec^{-1}x \). \[ y = \cot^{-1}\left(\frac{1}{\sqrt{\sec^2\theta - 1}}\right) \] Using the identity \( \sec^2\theta - 1 = \tan^2\theta \): \[ y = \cot^{-1}\left(\frac{1}{\sqrt{\tan^2\theta}}\right) \] Since \( x>1 \), we have \( \theta \in (0, \pi/2) \). In this interval, \( \tan\theta \) is positive. Therefore, \( \sqrt{\tan^2\theta} = \tan\theta \). \[ y = \cot^{-1}\left(\frac{1}{\tan\theta}\right) \] We know that \( \frac{1}{\tan\theta} = \cot\theta \). \[ y = \cot^{-1}(\cot\theta) \] Since \( \theta \in (0, \pi/2) \), which is within the principal value branch of \( \cot^{-1} \), we can write: \[ y = \theta \] Now, substitute back the value of \( \theta \) in terms of \( x \). \[ y = \sec^{-1}x \] Step 4: Final Answer:
The simplest form of the expression is \( \sec^{-1}x \).