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If a + b + c = 180 degree prove that cos square A + cos square B + cos square C + 2 cos A ×cosB× cosC= 1
Given A+B+C= 180Taking LHS, Cos^2A+cos^2 B + cos^2 C+ 2cosA cosB cosC =(1+cos2a/2)+ (1+cos2b/2)+ (1+cos2c/2)+2cosa cosb cosc =3/2+1/2(cos2a + cos2b + cos2c ) + 2cosa cosb cosc ........(1)Take, cos2a+cos2b+cos2c =2 (cos 2a+2b\2)(cos2a-2b\2) + cos2c=2cos( a+b) cos (a-b)+cos2c=2cos(180-c) cos (a-b)+cos 2c=-2cosc cos(a-b)+2cos^2C - 1-1-2cosc[ cos( a-b)-cosc]=-1-2cosc[cos(a-b)-cos (180-(a+b))]=-1-2cosc[cos(a-b)+cos(a+b)]=-1-2cosc(2cosa cosb)=-1-4cosa cosb coscPut this value in eq..(1)LHS ,3\2+1\2(-1-4cosa cosb cosc) + 2 cosa cosb cosc =3\2-1\2-2cosa cosb cosc + 2 cosa cosb cosc =3\2-1\2=2\2=1 ....RHS LHS=RHS
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