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Question

In triangle ABC,tanA2,tanB2,tanC2 are in H.P., then the value of cotA2×cotC2 is equal to

A
1
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B
2
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C
3
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D
4
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Solution

The correct option is C 3
tanA/2,tanB/2,tanC/2 are in HP
1tanA/2,1tanB/2,1tanC/2 are in A.P
2tanB/2=1tanA/2+1tanC/2
2cotB/2=cotA/2+cotC/2
2cot(π(A+C)2)=cotA/2+cotC/2
2tan(A+C2)=cotA/2+cotC/2
2[tanA/2+tanC/21tanA/2tanC/2]=tanA/2+tanC/2tanA/2tanC/2
let tanA/2tanC/2=y
21y=1y2y=1y
y=1/3
cotA/2cotC/2=1y=3(C)

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