If A + B + C = 180°, Prove that

Question

Cos²A + Cos²B + Cos²C = 1-2cosAcosBcosC

Answer 1

2cosAcosBcosC =(2cosAcosB)cosC
=(cos(A+B) + cos(A-B))cosC
=cos(pi-C)cosC + cos(A-B)cos(pi-(A+B))
=-2cos^2C - 2cos(A+B)cos(A-B)
=-cos^2C - cos^2A + sin^2B

so,

1 - 2cosAcosBcosC = 1 + cos^2C + cos^2A - sin^2B
= 1 - sin^2B + cos^2C + cos^2A
= cos^2B + cos^2C + cos^2A

Answer 2

Much samaj mai nhi aaya