If A+B+C =180^@, prove that cot^2 A + cot^2 B +cot^2 C >= 1 ?

1 Answer
Oct 1, 2017

A+B+C=180^@

=>cot(A+B)=cot(180^@-C)

=>(cotAcotB-1)/(cotB+cotA)=-cotC

=>(cotAcotB-1)=-(cotB+cotA)*cotC

=>(cotAcotB+cotBcotC+cotCcotA=1

Now

cot^2A+cot^2B+cot^2C-1

=cot^2A+cot^2B+cot^2C-(cotAcotB+cotBcotC+cotCcotA)

=1/2(2cot^2A+2cot^2B+2cot^2C-2cotAcotB-2cotBcotC-2cotCcotA)

=1/2[(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

cot^2A+cot^2B+cot^2C-1>=0

=>cot^2A+cot^2B+cot^2C>=1