If the angles A, B, C of a triangle are in A.P., then acsin⁡2C+casin⁡2A=

If the angles A, B, C of a triangle are in A.P., then acsin2C+casin2A=

  1. A

    12

  2. B

    32

  3. C

    1

  4. D

    3

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    Solution:

    A,B,C are in  A.P ThenB=60

    acsin2C+casin2A=sinAsinC2sinCcosC+sinCsinA2sinAcosA

    =2(sinAcosC+cosAsinC) =2sin(A+C)=2sin120=3

     

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