Question

cs 2A-cos 2B-cos2C=1-4sinA cosB sinC

Answer 1

L.H.S. = cos 2A – cos2B + cos2C = -2 sin(A + B) . sin(A – B) + 1 – 2sin2C = 1 – 2 sinC sin (A – B) – 2 sin2C [∵ sin(A+B)-sinc] = 1 – 2 sinC [sin(A – B) + sinC] = 1 – 2 sinC [sin(A + B) + sin(A – B)] = 1 – 2 sinC [2sinA . cosB] = 1 – 4 sinA cosB sinC = R.H.S.