If A + B + C = π then Prove that: cos²A + cos²B – cos²C+ 2sinA.sinB.cosC=1

Solution: Here,A + B + C = π then B+C=π-A

Now, Let’s consider

2 sinA . sinB . cosC
= sinA [2 sinB.cosC]
= sinA [sin(B+C) + sin(B-C)] =sinA[sin(180-A) + sin(B-C)]
= sinA [sinA + sin(B-C)]
= sin²A + sinA sin(B-C)
= sin²A + {2sinA sin(B-C) }/ 2
= sin²A + {cos(A-B+C) – cos(A+B-C) }/ 2
= sin²A + {cos(A+C-B) – cos(A+B-C) }/ 2
= sin²A + {cos(180-B-B) – cos(180-C-C)}/2
= sin²A + {cos(180-2B) – cos(180-2C)}/ 2
= sin²A – {cos2B + cos2C }/ 2
= ( 2sin²A – cos2B + cos2C ) / 2
= 1/2 { 2sin²A – cos2B + cos2C }
= 1/2 { 2sin²A – 1+2sin²B + 1-2sin²C }
=1/2 { 2sin²A + 2sin²B – 2sin²C }
= sin²A + sin²B – sin²C
= sin²A + sin²B – sin²C = 2 sinA . sinB . cosC
= 1 – cos²A + 1-cos²B -1+cos²C =2sinA.sinB.cosC
= cos²A + cos²B – cos²C
= 1- 2sinA.sinB.cosC

Hence Proved