Question

sin2A-Sin2B+Sin2C=4CosAsinBcosC​

Answer 1

Step-by-step explanation:

Given A + B + C = 180°

2A + 2B + 2C = 360°

2A + 2B = 360° – 2C

sin(2A + 2B) = sin(360° – 20) = – sin2C

cos(2A + 2B) = cos(360° – 2C) = cos 2C

L.H.S = sin2A – sin2B + sin2C

= 2cos(A + B) · sin(A – B) + 2sinC. cosC

= – 2cosC.sin(A – B) + 2 sinc.cosC

= 2 cosC [sinC – sin (A – B)]

= 2 cosC [sin(A + B) – sin(A – B)]

= 2 cos C [2cosA . sinB]

= 4 cosA sinB.cosC = R.H.S.

Hence Proved //

Hope it Helps you !