If θ1,θ2,θ3,…,θn are in AP, whose common difference is d, then  sin⁡ dsec⁡θ1sec⁡θ2+sec⁡θ2sec⁡θ3+…+sec⁡θn−1sec⁡θn is equal to

If θ1,θ2,θ3,,θn are in AP, whose common difference is d, then  sin dsecθ1secθ2+secθ2secθ3++secθn1secθn is equal to

  1. A

    tanθntanθ2

  2. B

    tanθn+tanθ1

  3. C

    tanθntanθ1

  4. D

    None of these

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    Solution:

    Since,θ1,θ2,θ3,,θn are in AP

     θ2θ1=θ3θ2==θnθn1=d      (i)

    Now, taking only first term

    sindsecθ1secθ2=sindcosθ1cosθ2=sinθ2θ1cosθ1cosθ2=sinθ2cosθ1cosθ2sinθ1cosθ1cosθ2=sinθ2cosθ1cosθ1cosθ2cosθ2sinθ1cosθ1cosθ2=tanθ2tanθ1

    Similarly, we can solve other terms which will be

    tanθ3tanθ2,tanθ4tanθ3,sindsecθ1secθ2+secθ2secθ3++secθn1secθn=tanθ2tanθ1+tanθ3tanθ2 ++tanθntanθn1 =tanθ1+tanθn=tanθntanθ1 

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