Question

If in triangle ABC , sin A sin C = sin ( A- B ) sin ( B - C ) prove that a 2 , b 2 , c 2 are in A.P

Answer 1

Answer:

sinA=sin(B−C)sin(A−B)

sin(π−(A+B)sin(π−(B+C)=sin(B−C)sin(A−B)

sin(A+B)sin(B+C)=sin(B−C)sin(A−B)

2sin(B+C)sin(B−C)=2sin(A+B)sin(A−B)

cos2C−cos2B=cos2B−cos2A

2cos2B=cos2A+cos2C

2(1−2sin2B)=2−2sin2A−2sin2C

1−2sin2B=1−sin2A−sin2C

2sin2B=sin2A+sin2C

Hence,

sin2A,sin2B,sin2C are in A.P.

Hence, by sine rule a2,b2,c

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