Question

cos²a+cos²b-cos²c=1-2sina.sinb.sinc​

Answer 1

Answer:

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