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Question

If θ=tan1d1+a1a2+tan1d1+a2a3++tan1d1+an1an, where a1,a2,a3,an  are in A.P. with common difference d, then tanθ=

 
  1. (n1)da1+an
  2. (n1)d1+a1an
  3. ana1an+a1
  4. nd1+a1an


Solution

The correct option is B (n1)d1+a1an
θ=tan1a2a11+a1a2+tan1a3a21+a2a3++tan1anan11+an1an
=(tan1a2tan1a1)+(tan1a3tan1a2)++tan1anan11+an1an++(tan1antan1an1)
=tan1antan1a1
=tan1ana11+a1an=tan1(n1)d1+a1an
tanθ=(n1)d1+a1an

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