Question

Cos2a+cos2b+cos2(a-b) +1=4cosa.cosb.cos(a-b)

Answer 1

Answer:

L.H.S. = cos2A + cos2B + cos2C

= 2cos(A + B) · cos(A – B) + 2cos2C – 1

= -2cosC . cos(A – B) + 2cos2C – 1

= -2cosC[cos (A – B) – cos C] – 1

= -2cosC{cos (A – B) + cos(A + B)} – 1

= -2 cosC (2cosA · cosB) – 1

= – 1 – 4 cos A cosB cosC = R.H.S.