We need to evaluate the expression \( \cos \left[ \cot^{-1} \left( -\sqrt{3} \right) + \frac{\pi}{6} \right] \). \textbf{Step 1: Understanding} \( \cot^{-1}(-\sqrt{3}) \): The function \( \cot^{-1}(x) \) gives the angle whose cotangent is \( x \). We need to find the angle \( \theta \) such that: \[ \cot(\theta) = -\sqrt{3} \] The angle \( \theta \) where \( \cot(\theta) = -\sqrt{3} \) is \( \frac{5\pi}{6} \), because \( \cot \left( \frac{5\pi}{6} \right) = -\sqrt{3} \). Thus, \[ \cot^{-1}(-\sqrt{3}) = \frac{5\pi}{6} \] \textbf{Step 2: Substituting into the original expression:} Now, substitute this value into the original expression: \[ \cos \left( \frac{5\pi}{6} + \frac{\pi}{6} \right) \] \textbf{Step 3: Simplifying the angle:} To simplify, add the angles: \[ \frac{5\pi}{6} + \frac{\pi}{6} = \frac{6\pi}{6} = \pi \] \textbf{Step 4: Finding \( \cos(\pi) \):} From trigonometric values, we know that: \[ \cos(\pi) = -1 \] Thus, the correct answer is (D) -1.
If \( \alpha>\beta>\gamma>0 \), then the expression \[ \cot^{-1} \beta + \left( \frac{1 + \beta^2}{\alpha - \beta} \right) + \cot^{-1} \gamma + \left( \frac{1 + \gamma^2}{\beta - \gamma} \right) + \cot^{-1} \alpha + \left( \frac{1 + \alpha^2}{\gamma - \alpha} \right) \] is equal to: