Solution:
Statement 1: In a triangle ABC, if cotA⋅cotB⋅cotC>0, then the triangle is an acute-angled triangle.
An acute-angled triangle is a triangle where all angles are less than 90∘. The cotangent function cotθ is positive for 0∘<θ<90∘. Therefore, for all angles A,B, and C in an acute-angled triangle, we have:
0∘<A,B,C<90∘
⟹cotA>0,cotB>0,cotC>0
Thus, the product cotA⋅cotB⋅cotC will indeed be greater than zero. This means statement 1 is true: If cotA⋅cotB⋅cotC>0, then the triangle is acute-angled.
Statement 2: In a triangle ABC, if tanA⋅tanB⋅tanC>0, then the triangle is an obtuse-angled triangle.
Given that the sum of the angles in a triangle is 180∘, i.e., A+B+C=180∘, we also know that the product of the tangents of the angles in any triangle is given by:
tanA⋅tanB⋅tanC=tanA⋅tanB⋅tan(180∘−A−B)=tanA⋅tanB⋅(−tan(A+B))
=−tanA⋅tanB⋅tan(A+B)
Since tan(A+B)=−tanC, the above product simplifies to:
tanA⋅tanB⋅tanC=−tanA⋅tanB⋅tan(180∘−A−B)<0
Therefore, tanA⋅tanB⋅tanC will be negative in general triangles and not positive. If tanA⋅tanB⋅tanC>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:
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