If a=2sin⁡θ1+cos⁡θ+sin⁡θ, then 1+sin⁡θ−cos⁡θ1+sin⁡θ is equal to 

If a=2sinθ1+cosθ+sinθ, then 1+sinθcosθ1+sinθ is equal to 

  1. A

    1a

  2. B

    a

  3. C

    1-a

  4. D

    1+a

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    Solution:

    We have,

      a=2sinθ1+cosθ+sinθ

     a=2sinθ(1+sinθcosθ)(1+cosθ+sinθ)(1+sinθcosθ) a=2sinθ(1+sinθcosθ)(1+sinθ)2cos2θ a=2sinθ(1+sinθcosθ)2sinθ(1+sinθ)=1+sinθcosθ1+sinθ

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