If A+B+C=3π2,then cos2A+cos⁡2B+cos⁡2C is equal to

If A+B+C=3π2,then cos2A+cos2B+cos2C is equal to

  1. A

    14cosAcosBcosC

  2. B

    4sinAsinBsinC

  3. C

    1+2cosAcosBcosC

  4. D

    14sinAsinBsinC

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    Solution:

    Since,A+B+C=3π2

    cos2    A+cos2B+cos2C    =2cos(A+B)cos(AB)+cos2C    =2cos3π2Ccos(AB)+12sin2C    =12sinCcos(AB)+sin3π2(A+B)    =12sinC[cos(AB)cos(A+B)]    =14sinAsinBsinC

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