Prove that cotA + cotB + cotC = 0, when A, B and C are angles of a triangle.

Question

Prove that cotA + cotB + cotC >= 0, when A, B and C are angles of a triangle.

Subhashis · Answer

Since value of cot theta lies between 0and infinity...therefore..if we take the most minimum values of a b and c functioned as cot then also it would be greater than zero...i.e,0+0+0(cot a+cot b+cot c)=0...pd

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Prove that cotA + cotB + cotC >= 0, when A, B and C are angles of a triangle.

Prove that cotA + cotB + cotC >= 0, when A, B and C are angles of a triangle.

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