Let A=θ∈−π2,π:3+2isin⁡θ1−2isin⁡θ is purely imaginary . Then the sum of the elements in A .is

Let A=θπ2,π:3+2isinθ12isinθ is purely imaginary . Then the sum of the elements in A .is

  1. A

    3π/4

  2. B

    2π/3

  3. C

    π

  4. D

    5π/6

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    Solution:

    Let z=3+2isinθ12isinθ

    =3+2isinθ12isinθ×1+2isinθ1+2isinθ=34sin2θ+i(8sinθ)1+4sin2θ

    Now, (z)=0                                        [ z is purely imaginary]

    34sin2θ1+4sin2θ=0sin2θ=34sin2θ=sin2π3 θ=nπ±π3,nZ

    θ=π3,π3,2π3 θπ2,π

    Hence, sum of elements of fA=π3+π3+2π3=2π3

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