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Question

If A, B, C are acute positive angles such that A+B+C=π and cotA cotB cotC=K, then

A
K133
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B
K133
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C
K<19
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D
K>13
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Solution

The correct option is A K133

We have,

tan(A+B+C)=(tanA+tanB+tanCtanAtanBtanC)(1tanAtanBtanBtanCtanCtanA)


A+B+C=π


tan(A+B+C)=tanπ=0


tanA+tanB+tanC=tanAtanBtanC ....(1)
As all angles are acute, we can apply the A.M.G.M. inequality for tanA,tanB,tanC


tanA+tanB+tanC3(tanAtanBtanC)1/3

Using equation (1), and cotAcotBcotC=K


13K(K)1/3


K133


cotAcotBcotC133



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