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manoj jangra Grade: 12
```         If A+B+C = ∏ and A,B,C belongs to (0,∏/2) then prove that
secA +secB +secC≥6```
7 years ago

Answers : (1)

suryakanth AskiitiansExpert-IITB
105 Points
```										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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Suryakanth –IITB

```
7 years ago
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