1.

If A+B+C =(3pi)/(2), then cos2A+cos2B+cos2C is equal to-

Answer»

`1-4cosAcosBcosC`
`4sinAsinBsinC`
`1+2cosA+cosBcosC`
`1-4sinAsinBsinC`

Solution :`cos2A+cos2B+cos2C=2COS(A+B)COS(A-B)+cos2C`
`=2cos((3pi)/2-C) cos(A-B)+cos2C THEREFORE A+B+C=(3pi)/(2)`
`=-2sinCcos(A-B)+1-2sin^(2)C=1-2sinC[cos(A-B)+sinC]`
`=1-2sinC[cos(A-B)+SIN((3pi)/(2)-(A+B))]`
`=1-2sinC[cos(A-B)-cos(A+B)]=1-4sinBsinC`
`=1-2sinC[cos(A-B)+sin((3pi)/2-(A+B))]`
`=1-2sinC[cos(A-B)-cos(A+B)]=1-4sinAsinBsinC`


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