24I too ahve done this ques by that method only..so I dont know..but i will be intrested to see any new method
Prove that the determinant --
sin 2A sin C sin B
sin C sin2B sin A
sin B sin A sin2C
=0
where A,B,C are the angles of a triangle.
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6 Answers
24
1write sin2A = 2sinAcosA = sinAcosA + sinAcosA
sinC= sin[pi - (A+B)] = sin(A+B) = sinAcosB + sinBcosA
and so on for all the terms..and then write it as a product of two determinants...that way its really easy...
but i'm really lazy...so not feeling like typing all that long stuff!! [3]