If π2<θ<π, then 1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ is equal to is

If π2<θ<π, then 1sinθ1+sinθ+1+sinθ1sinθ is equal to is

  1. A

    2secθ

  2. B

    2secθ

  3. C

    sec θ

  4. D

    -secθ

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    Solution:

    We have,

    1sinθ1+sinθ+1+sinθ1sinθ=1sinθ+1+sinθ1sin2θ=2|cosθ|=2cosθ=2secθ,   π/2<θ<πcosθ<0

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