Question

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

Answer 1

`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`