Math, asked by deepu190822, 1 year ago

sin2a-sin2b-sin2c÷sin2b-sin2a-sin2csin 2A - sin 2B - sin 2c by sin 2B - sin 2A - sin 2c ​

Answers

Answered by nikolatesla2
1

Answer:

Step-by-step explanation:

LHS = sin 2A + sin 2B + sin 2C = sin (pie – 2A) + sin (pie –2B) + sin (pie –2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A - B + C)/2 cos (A - B - C)/2) = 2sinC (2cos (pie - 2B)/2 cos(2A - pie)/2 = 4sinAsinBsinC = RHS Hence Proved

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