Question

Question #0e494

Answer 1

For `DeltaABC`

we have `A+B+C=pi`

So `cot(A+B)=cot(pi-C)`

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

`=>cotAcotB+cotBcotC+cotCcotA=1" "color(red)([1])`

Now it is given

`cotA+cotB+cotC=sqrt3`

Squaring both sides we get

`cot^2A+cot^2B+cot^2C+2(cotAcotB+cotBcotC+cotCcotA)=3" "color(red)([2])`

Now subtracting thrice of [1] from [2] we get

`cot^2A+cot^2B+cot^2C-(cotAcotB+cotBcotC+cotCcotA)=0`

`=>2cot^2A+2cot^2B+2cot^2C-2cotAcotB-2cotBcotC-2cotCcotA=0`

`=>(cotA-cotB)^2+(cotB-cotC)^2+(cotC-cotA)^2=0`

Now sum of three squared terms each of which is positive being zero each will be equal to zero.

Hence

`cotA=cotB=cotC`

`=>A=B=C`

This implies that `DeltaABC` is equilateral.