Question

A+B+C=1 prove that cotAcotC+cotC. CotA+cotA. CotB=2

Answer 1

Answer:

In ΔABC , A+B+C=180

∘

⟹A+B=180

∘

−C

⟹cot(A+B)=cot(180

∘

−C)

⟹

cotA+cotB

cotA.cotB−1

=−cotC

⟹cotA.cotB−1=−cotA.cotC−cotB.cotC

⟹cotA.cotB+cotB.cotC+cotC.cotA=1