Question

sin2A-sin2B+sin2C=4cosAsinBcosc please solve this problem ​

Answer 1

Answer:

a+b=180-c

L.H.S=(SIN2A-SIN2B)+SIN2C

        =2COS(A+B)SIN(A-B)+2SINC.COSC

       =-2COS C SIN(A-B)+2 COS C SIN(A+B)

       =2COS C[SIN(A+B)-SIN(A-B)]

       =2COS C[2SIN B COS A]

       =4 cosA sinB cosC