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Question

If ABC is a triangle and tanA2,tanB2,tanC2 are in H.P., then find the minimum value of cotB/2.

A
3+1
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B
3
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C
2
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D
2+1
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Solution

The correct option is B 3
As tanA2,tanB2,tanC2H.P
tanB2=2tanA2tanC2tanA2+tanC2
=2sinA2cosA2.sinC2cosC2sinA2cosA2+sinC2cosC2=2sinA2sinC2sin(A2+C2)
=cos(A2C2)cos(A2+C2)sin(A2+C2)
=cos(A2+C2)cos(π2+B2)sin(π2B2)
=cos(A2C2)sinB2cosB2
sinB2=cos(A2C2)sinB2
2sinB2=cos(A2C2)
cotB2<3

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