If cos⁡α+cos⁡β=0=sin⁡α+sin⁡β then cos⁡2α+cos⁡2β is equal to

If cosα+cosβ=0=sinα+sinβ then cos2α+cos2β is equal to

  1. A

    2cos(α+β)

  2. B

    -2cos(α+β)

  3. C

    3cos(α+β)

  4. D

    None of these

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    Solution:

    Given cosα+cosβ=0 and sinα+sinβ=0

    On squaring and subtracting both the equations, we get

     (cosα+cosβ)2(sinα+sinβ)2=0cos2α+cos2β+2cosαcosβsin2α+sin2β+2sinαsinβ=0cos2αsin2α+cos2βsin2β+2[cosαcosβsinαsinβ]=0cos2α+cos2β+2cos(α+β)=0cos2α+cos2β=2cos(α+β)

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