`cos^2 A + cos^2 B +cos^2 C=1+2cosA cosB cosC`.

Explanation

(a) `LHS=cos^(2)A+cos^(2)A+cos^(2)B+cos^(2)C` ltbr. `=(3+(cos 2A+cos2B+cos2C))/(2)`
now given `A+B=C`
or `A+B+(pi-C)=pi`
or `A+B+D=pi`, where `D=pi-C`
`rArr LHS=(3+(cos2A+cos2B+cos2C))/(2)`
`=(3+(Cos2A+Cos2B+cos2D))/(2)`
`=1-2cos A cos B cos D`
`=1-2cos A cos B cos (pi-C)`
`=1+2cos A cos B cos C=RHS`.
(b) Given `alpha+beta=60^(@)`
`cos^(2)alpha+cos^(2)beta-cos alpha cos beta`
`=1-sin^(2)alpha+cos^(2)beta-cos alpha cos beta`
`=1+cos(alpha+beta)cos(alpha-beta)-cos alpha cos beta`
`=1+cos60^(@)cos(alpha-beta)-cos alpha cos beta`
`=1+(cos alpha cos beta+sin alpha sin beta)/(2)-cos alpha cos beta`
`=1-(cos(alpha+beta))/(2)`
`=1-(1)/(4)=(3)/(4)`