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Question

If in a ABC, tanA2,tanB2,tanC2 are in H.P., then the minimum value of cotB2 is
  1. 3
  2. 2
  3. 3
  4. 2


Solution

The correct option is C 3
In a ABC,
cotA2+cotB2+cotC2=cotA2cotB2cotC2     (1)

tanA2,tanB2,tanC2 are in H.P.
cotA2,cotB2,cotC2 are in A.P.
2cotB2=cotA2+cotC2
From (1),
3cotB2=cotA2cotB2cotC2cotA2cotC2=3

Now, A.M.G.M. 
cotA2+cotC22cotA2cotC22cotB223cotB23

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