If A+B+C=180∘, prove that cot2A+cot2B+cot2C≥1 ?

1 Answer
P dilip_k
Oct 1, 2017

A+B+C=180∘

⇒cot(A+B)=cot(180∘−C)

⇒cotAcotB−1cotB+cotA=−cotC

⇒(cotAcotB−1)=−(cotB+cotA)⋅cotC

⇒(cotAcotB+cotBcotC+cotCcotA=1

Now

cot2A+cot2B+cot2C−1

=cot2A+cot2B+cot2C−(cotAcotB+cotBcotC+cotCcotA)

=12(2cot2A+2cot2B+2cot2C−2cotAcotB−2cotBcotC−2cotCcotA)

=12[(cotA−cotB)2+(cotB−cotC)2+(cotC−cotA)2]
This being sum of three squared quantities each of which is difference of two quantities. So this sum is ≥0

Hence

cot2A+cot2B+cot2C−1≥0

⇒cot2A+cot2B+cot2C≥1

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