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If A+B+C = ∏ and A,B,C belongs to (0,∏/2) then prove that
secA +secB +secC≥6
Dear manoj,
A,B,C are less than 90 degrees hence
cosA>cosB>cosC (consider A<B<C)
We have, secA<secB<secC
Here,we assume all of them to be positive,otherwise -infinity will be the minimum value.
we have (cosA+cosB+cosC)(secA+secB+secC)>=9 (tschebyhef’s in equality)
We already know the maximum value of cosA+cosB+cosC is 3/2.(we can prove it easily)
=> (secA+secB+secC)>= 9*(2/3)
Hence
(secA+secB+secC)>=6
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