Question

sin²2A+sin²2B+sin²2C=2(1-cos2Acos2Bcos2c)​

Answer 1

Answer:

hey mate here is your answer

Step-by-step explanation:

L.H.S. = cos2A + cos2B – cos2C = 1212[1+2cos(A+B) cos(A – B) – (2cos2C – 1)] = 1212[1+2cos(A+B).cos(A – B) – 2cos2C +1] = 1212[2 + 2(-cosc) cos(A – B) 2cos2C] = 1212[2 – 2 cos C[cos (A – B) + cos C] = 1212[2 – 2cosC[cos(A – B) – cos(A+B)]] = 1 – cosC[-2sin A . sin(-B)] =1 – cosC[2sin A sin B] = 1 – 2sinA sinB cosC = R.H.S.