Kerala KEAM 5 June 2024 Solved Paper

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).Quick Tip: 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.