UPSC NDA II 2024 Mathematics Paper

Concept:In any triangle, the sum of angles is $\pi$.Using the tangent addition formula for three angles leads to a relation between $\tan A \tan B \tan C$ and $\tan A + \tan B + \tan C$.Then, $\cot A \cot B \cot C$ is the reciprocal of that product.Explanation:In triangle $ABC$, $A + B + C = \pi$.Take tangent on both sides: $\tan(A + B + C) = \tan \pi = 0$.Expand using the tangent sum formula for three angles:$$\frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)} = 0$$The denominator is finite, so the numerator must be zero:$$\tan A + \tan B + \tan C - \tan A \tan B \tan C = 0$$Thus, $\tan A \tan B \tan C = \tan A + \tan B + \tan C = k$.Now, $\cot A \cot B \cot C = \frac{1}{\tan A \tan B \tan C} = \frac{1}{k}$.Answer:Option B: $\frac{1}{k}$