Question

Q.10

 

In a ΔABC, show thatsin²A + sin²B - sin²C = 2 sin A. sin B. cos C​

Answer 1

Answer:

Consider the problem,

sin2A+sin2B+sin2c=2+2cosA.cosB.cosc

We can write sin2A as,

sin2A=21−cos(2A)

Therefore,

LHS=21−cos(2A)+21−cos(2B)+21−cos(2C)=23−(cos(2A)+cos(2B)+cos(2C))=21(3−(2cos(A+B)cos(A−B)+cos(2C)))C=180−(A+B)cos(C)=cos(180−(A+B))cos(C)=−cos(A+B)

Therefore,

=23−(−2cos(C)cos(A−B)+cos(2C))cos(2C)=2

Answer 2

L.H.S. = sin2A + sin2B – sin2C = 1212[1-(cos 2A + cos2B – cos2C)] = 1212[1-{{2cos(A + B). Cos(A – B) – 2cos2C+1}] = 1212[+2cosC . cos(A – B) + 2cos2c] = 1212[2cosC(cos(A – B) + cosC] = 1212[2cosC(cos(A – B) – cos(A + B))] = 1212[2cosC[-2sin A.sin(-B)]] = 2 sinA sinB sin