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The incircle touches side BC of triangle ABC at D and ID is produced to H so that DH=s, where s and I are the semi-perimeter and incentre of triangle ABC respectively. If HBIC is cyclic, then cot(A4) is equal to 
  1. 23
  2. 2+1
  3. 21
  4. 2+3


Solution

The correct option is B 2+1

As HBIC is cyclic, so
rs=(sb)(sc)
=(sb)(sc)
12bcsinA=(sb)(sc)
12sinA=(sb)(sc)bc=sin2A2
sinA=2sin2A2
2sinA2cosA2=2sin2A2
sinA2(cosA2sinA2)=0
cotA2=1, so A2=π4
A4=π8
Hence, cotA4=cotπ8=2+1

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