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

Answer

1

Explanation

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`