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if A+B+C=pi and A,B,,C are acute angles then prove that tanA+tanB+tanC>=33/2

Nishtha Gahlot , 7 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 7 Years ago

Using A.M. > = G.M.
(tanA + tanB + tanC)/3 >= (tanA.tanB.tanC)^(1/3)
=> tanA + tanB + tanC >= 3*(tanA + tanB + tanC)^(1/3)...
(since in triangle sum of tangents of angles is equal to product of tangents)
=> tanA + tanB + tanC >= 3(3)^(1/2)
tanA + tanB + tanC = 3(3)^(1/2) implies
A.M. = G.M. and this means tanA = tanB = tanC
=> A =B =C and hence triangle is equilateral,
 
If we take triangle as equilateral;
then all the angles are 60 degree, then tanA+ tanB + tanC = 3(3)^(1/2)
 
Regards
Arun (askIITians forum expert)

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