Q2=In an Acute angled Triangle ABC,Prove That secA(1+secA)secB(1+secB)secC(1+secC)is >=216

Asked by  | 28th Oct, 2012, 11:47: AM

Expert Answer:

Minimum value of the given expression will occur when triangle will be equilateral so A=B=C=?/3

Minimum value of expression will be=27*8=216

secA(1+secA)secB(1+secB)secC(1+secC)is >=216

Answered by  | 18th Nov, 2012, 07:51: PM

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