If cos⁡x−cos⁡αcos⁡x−cos⁡β=sin2αcos⁡βsin2⁡βcos⁡α, then 

  If cosxcosαcosxcosβ=sin2αcosβsin2βcosα, then 

  1. A

    cosx=2cosα+cosβ1+cosαcosβ

  2. B

    cosx=cosa+cosβ1cosαcosβ

  3. C

    tanx2=tanα2tanβ2

  4. D

    tanx2=2tanα2tanβ2

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    Solution:

    From the given results 

    cosxsin2βcosαcos2αsin2β=sin2αcosβcosxsin2αcos2β cosxcosαsin2βsin2αcosβ=cos2αsin2βsin2αcos2β

     cosx=cos2α1cos2β1cos2αcos2βcosα1cos2β1cos2αcosβ=cos2αcos2β(cosαcosβ)(cosαcosβ+1)=cosα+cosβ1+cosαcosβcosx1=cosα+cosβ1+cosαcosβ1cosx1+cosx=1+cosαcosβcosαcosβ1+cosαcosβ+cosα+cosβ=(1cosα)(1cosβ)(1+cosα)(1+cosβ)

    1cosx1+cosx=1+cosαcosβcosαcosβ1+cosαcosβ+cosα+cosβ=(1cosα)(1cosβ)(1+cosα)(1+cosβ)tan2x2=tan2α2tan2β2tanx2=±tanα2tanβ2tanx2=±tanα2tanβ2

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