InterviewSolution
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InterviewSolution
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For any `triangleABC`, the value of determinant `|[sin^2A,cotA,1],[sin^2B,cotB,1],[sin^2C,cotC,1]|` is: |
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Answer» `Delta = |[sin^2A,cotA,1],[sin^2B,cotB,1],[sin^2C,cotC,1]|` `=>Delta = [sin^2A(cotB-cotC) -sin^2B(cotA - cotC)+ sin^2C( cotA - cotB)]` Converting, `cot` into `cos/sin` form, we get, `=>Delta = [sin^2A(sin(C-B)/(sinB*SinC)) +sin^2B(sin(A-C)/(sinC*SinA))+ sin^2C( sin(B-A)/(sinB*SinA))]` `=>Delta = [(sin^3A(sin(C-B))+sin^3B(sin(A-C))+ sin^3C( sin(B-A)))/(sinAsinBsinC)]` Now, `sin^3A = sinAsin^2A = sin(pi-(B+C))sin^2A = sin(B+C)sin^2A)` `sin^3B = sinBsin^2B = sin(pi-(A+C))sin^2B = sin(A+C)sin^2B)` `sin^3C = sinCsin^2C = sin(pi-(B+A))sin^2C = sin(B+A)sin^2C)` So, our equation becomes, `Delta = [ (sin^2A((sinCcosB)^2-(cosCsinB)^2)+sin^2B((sinAcosC)^2-(cosAsinC)^2)+ sin^2C( (sinBcosA)^2-(cosBsinA)^2))/(sinAsinBsinC)]` `Delta = [0/(sinAsinBsinC)] =0` |
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