If `A+B+C=270^@` , then `cos2A+cos2B+cos2C+4sinAsin B sinC=` – InterviewSolution

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1.	If `A+B+C=270^@` , then `cos2A+cos2B+cos2C+4sinAsin B sinC=`
Answer» Correct Answer - B We have, `cos2A+cos2B+cos2C+4sinAsinBsinC` `=2cos(A+B)cos(A-B)+cos2C+4sinA sinBsinC` `=2cos((3pi)/(2)-C)cos(A-B)+cos2C+4sinA sinBsinC` `=-2sinCcos(A-B)+1-2sin^(2)C+4sinAsinBsinC` `=-2sinC{cos(A-B)+sinC}+4sinAsinBsinC+1` `=-sinC{cos(A-B)-cos(A+B)}+4sinAsinBsinC+1` `=-4sinAsinBsinC+4sinA sinBsinC+1=1`

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