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

1 Answer
P dilip_k
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#

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