`a^2cos(2B)+b^2cos(@A)+2abcos(A-B)=c^2`

If A+B+C=pi then prove cos( (A)/2) cos( (B-C)/2) + cos( B/2) cos((C-A)/2) + cos( C/2) cos( (A-B)/2) = sinA +sinB+sinC

a^2(cos^2B-cos^2C)+b^2(cos^2C-cos^2A)+c^2(cos^2A-cos^2B)=0 .

In /_\ABC prove that a^2(cos^2B-cos^2C)+b^2(cos^2C-cos^2A)+c^2(cos^2A-cos^2B) = 0

a^(2)(cos^(2)B-cos^(2)C)+b^(2)(cos^(2)C-cos^(2)A)+c^(2)(cos^(2)A-cos^(2)B)=0

In any Delta ABC, prove that :a^(2)(cos^(2)B-cos^(2)C)+b^(2)(cos^(2)C-cos^(2)A)+c^(2)(cos^(2)A-cos^(2)B)=0

(iv) a^(2)(cos^(2) B - cos^(2)C) + b^(2) (cos^(2) C- cos^(2)A) + c^(2) (cos^(2)A- cos^(2)B)=0

If A+B+C=180^0 , prove that : cos^2(A/2) + cos^2(B/2) - cos^2(C/2) = 2cos(A/2) cos(B/2) sin( C/2)

If A+B+C=pi prove that cos^(2) A+cos^(2) B+cos^(2) C=1 - 2cos A cos B cos C .

If A+B+C=(3pi)/(2) , prove that cos ^(2)A+ cos ^(2) B- cos ^(2)C=-2 cos A cos B sin C.