UPSC NDA I 2024 Mathematics Paper

Statement 1: In a triangle $A B C$, if $\cot A \cdot \cot B \cdot \cot C>0$, then the triangle is an acute-angled triangle.An acute-angled triangle is a triangle where all angles are less than $90^{\circ}$. The cotangent function $\cot \theta$ is positive for $0^{\circ} < \theta < 90^{\circ}$. Therefore, for all angles $A, B$, and $C$ in an acute-angled triangle, we have:$0^{\circ} < A, B, C < 90^{\circ}$$\Rightarrow \cot A>0, \cot B>0, \cot C>0$Thus, the product $\cot A \cdot \cot B \cdot \cot C$ will indeed be greater than zero. This means statement 1 is true: If $\cot A \cdot \cot B \cdot \cot C>0$, then the triangle is acute-angled.Statement 2: In a triangle $A B C$, if $\tan A \cdot \tan B \cdot \tan C>0$, then the triangle is an obtuse-angled triangle.Given that the sum of the angles in a triangle is $180^{\circ}$, i.e., $A+B+C=180^{\circ}$, we also know that the product of the tangents of the angles in any triangle is given by:$\tan A \cdot \tan B \cdot \tan C = \tan A \cdot \tan B \cdot \tan (180^{\circ} - A - B) = \tan A \cdot \tan B \cdot (-\tan(A+B))$$= -\tan A \cdot \tan B \cdot \tan (A+B)$Since $\tan (A+B) = -\tan C$, the above product simplifies to:$\tan A \cdot \tan B \cdot \tan C = -\tan A \cdot \tan B \cdot \tan (180^{\circ} - A - B) < 0$Therefore, $\tan A \cdot \tan B \cdot \tan C$ will be negative in general triangles and not positive. If $\tan A \cdot \tan B \cdot \tan C>0$, this statement does not hold for obtuse triangles specifically but would imply an error. Thus, statement 2 is false.Based on the analysis above, we conclude that: