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min value of secA + secB + secC is............

min value of


secA + secB + secC is............


 

Grade:12

1 Answers

Askiitians_Expert Yagyadutt
askIITians Faculty 74 Points
14 years ago

Hii Piyush

 

Look  in case of triangle...

 

cosA + cosB + cosC >= 3/2    ( as A + B + C = 180 )

 

Now apply AM>GM   =>   (cosA + cosB + cosC)/3  >=  ( cosA.cosB.cosC)^1/3

 

1/2 >= (cosA.cosB.cosC)^1/3     =>   secA.secB.secC >= 8

 

Now Again apply AM>GM

 

(secA + secB + secC)/3 >= (secA.secB.secC)^1/3

 

so for minimum value of  secA + secB + secC  ..put secA.secB.secC = minimum i.e = 8

 

So...secA + secB + secC =  3 X (8)^1/3 = 6  ans

 

 

I hope you get it

 

Regards

Yagya

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