[A+B+C=180^(@)rArr cos2A+cos2B+cos2C=],[1)1-4sin A sin B sin C]

A+B+C=180^(@),cos2A+cos2B+cos2C=

cos2B + cos2A-cos2C = 1-4sin A sin B cos C

If A+B+C=90^(@) then (cos2A+cos2B+cos2C-1)/(sin A sin B sin C)=

If A+B+C=270^(@), then cos2A+cos2B+cos2C+4sin A sin B sin C=

cos2a-cos2b + cos2c = 1-4sin a cos b sin c

If A+B+C=180^(@) then cos2A+cos2B+cos2C=1-4cos A cos B cos C 2.If A+B+C=0 then cos A+cos B+cos C=-1+4(cos A)/(2)(cos B)/(2)(cos C)/(2)

A+B+C=180^(@)rArr SecA[cos B cos C-sin B sin C]

If A+B+C=270^(@) then cos2A+cos2B+cos2C=

If A+B+C=180^(@) , then (cos2A+cos2B+cos2C+1)/(cos A cos B cos C)

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