If A+B+C=π then,Prove that: cos²A + cos²B + cos²C + 2 cosA . cosB . cosC = 1

Solution: Here,A + B + C = π then

A+B=π-C

B+C=π-A

A+C=π-C

Now, Let’s consider

2 cosA . cosB . cosC
= cosA [2 cosB.cosC]

=cosA[cos(B+C)+ cos(B- C)]

=cosA [cos(180-A) + cos(B-C)]

=cosA [-cosA + cos(B-C)]

= -cos²A + cosA cos(B-C)
= -cos²A + {2cosA cos(B-C)}/2

=-cos²A + {cos(A-B+C) + cos(A+B-C)}/ 2
= -cos²A + {cos(A+C-B) + cos(A+B-C) }/ 2
= -cos²A + {cos(180-B-B) + cos(180-C-C)}/2
= -cos²A + {cos(180-2B) + cos(180-2C)}/2
= -cos²A – {cos2B + cos2C }/2
= ( -2cos²A – cos2B – cos2C ) / 2
= 1/2 { -2cos²A – cos2B – cos2C }
= 1/2 { -2cos²A – 2cos²B + 1 – 2cos²C +1}
=1/2 { -2cos²A – 2cos²B – 2cos²C + 2 }
= -(cos²A + cos²B + cos²C) + 1
or 2 cosA . cosB . cosC= – (cos²A + cos²B + cos²C) + 1

or cos²A + cos²B + cos²C + 2 cosA . cosB . cosC = 1

Hence Proved