Question #0e494

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Answer 1 · 2016-11-23T15:21:26

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.