Question

If \( x \neq 0, y \neq 0 \), then the value of \[ \cot^{-1}\left(\frac{x}{y}\right) + \cot^{-1}\left(\frac{y}{x}\right) \] is:

• A. \( \frac{\pi}{2} \)
• B. \( \frac{\pi}{2} \)
• C. 0
• D. \( -\pi \)
• E. \( -\frac{\pi}{2} \)

Hint:

The sum of two inverse cotangents can be simplified using the identity: \[ \cot^{-1}(a) + \cot^{-1}(b) = \cot^{-1}\left(\frac{ab - 1}{a + b}\right). \] This identity is helpful when simplifying expressions involving inverse trigonometric functions.

Answer 1

The Correct Option is B

We are given the expression: \[ \cot^{-1}\left(\frac{x}{y}\right) + \cot^{-1}\left(\frac{y}{x}\right). \] Using the identity for the sum of inverse cotangents: \[ \cot^{-1}(a) + \cot^{-1}(b) = \cot^{-1}\left(\frac{ab - 1}{a + b}\right), \] we substitute \( a = \frac{x}{y} \) and \( b = \frac{y}{x} \), and we get: \[ \cot^{-1}\left(\frac{x}{y}\right) + \cot^{-1}\left(\frac{y}{x}\right) = \cot^{-1}\left(\frac{\frac{x}{y} \cdot \frac{y}{x} - 1}{\frac{x}{y} + \frac{y}{x}}\right). \] Simplifying the expression: \[ \frac{x}{y} \cdot \frac{y}{x} = 1, \] so the numerator becomes: \[ 1 - 1 = 0, \] and the denominator becomes: \[ \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy}. \] Thus, the expression simplifies to: \[ \cot^{-1}(0) = \frac{\pi}{2}. \]
Therefore, the value of \( \cot^{-1}\left(\frac{x}{y}\right) + \cot^{-1}\left(\frac{y}{x}\right) \) is \( \frac{\pi}{2} \). Thus, the correct answer is option (B).