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If A+B+C=0, then prove that Det[[1,cosC,...

If A+B+C=0, then prove that `Det[[1,cosC,cosB],[cosC,1,cosA],[cosB,cosA,1]]=0`

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The correct Answer is:
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We have to prove `|(1,cosC,cosB),(cosC,1,cosA),(cosb,cosA,1)|=0`
`:.LHS=|(1,cosC,cosB),(cosC,1,cosA),(cosB,cosA,1)|`
`=1(1-cos^(2)A)-cosC(cosC-cosA.cosB)+cosB(cosC.cosA-cosB)`
`=sin^(2)-cos^(2)C+cosA.cosB.cosC+cosA.cosB.cosC-cos^(2)B`
`=sin^(2)A-cos^(2)B+2cosA.cosB.cosC-cos^(2)C`
`=-cos(A+B).cos(A-B)+2cosA.cosB.cosC-cos^(2)C`
`[ :'cos^(2)B-sin^(2)A=cos(A+B).cos(A-B)]`
`=-cos(-C).cos(A-B)+cosC(2cosA.cosB-cosC) [ :' cos (-theta)=costheta]`
`=-cosC(cosA.cosB+sinA.sinB-2cosA.cosB+cosC)`
`=cosC(cosA.cosB-sinA.sinB-cosC)`
`=cosC(cosC-cosC)=0=RHS` Hence proved
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