If cosα+cosβ=m and sinα+sinβ=n, then sin(α+β)=

If cosα+cosβ=m and sinα+sinβ=n, then sin(α+β)=

  1. A

    m2-n2m2+n2

  2. B

    n2-m2m2+n2

  3. C

    2mnm2+n2

  4. D

    2mnm2-n2

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    Solution:

    m=cosα+cosβ=2cosα+β2cosα-β2

    n=sinα+sinβ=2sinα+β2cosα-β2

     nm=tanα+β2 
    Now,

    sinα+β=2tanα+β21+tan2α+β2=2nm1+n2m2=2mnm2+n2


     

     

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