Question

PROVE THAT :
(sin2A+ sin2B + sin2C)/(sinA - sinB + sinC) = 8*cos A/2 *sinB/2 * cos c/2​

Answer 1

Answer:

HS = 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