Question #0e494

1 Answer
Nov 23, 2016

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.