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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`

Text Solution

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`=1[1-cos^2A]-cosc[cosc-cosAcosB]+cosB[cosAcosB-cosB]`
`=1-cos^2A-cos^2C+cosAcosBcosC+cosAcosBcosC-cos^2B`
`=1-[cos^A+cos^2B+cos^2C]+2coaAcosBcosC`
`=1-[cos^2A+cos^2B+(cos(A+B)^2)+2cosAcosB[A+B]]`
`=1-cos^2A-cos^2B-cos^2Acos^2B-acos^2b-2cosAcosBsin^2Asin^2B+2cosAcosBsinAsinB+2cos^2Acos^2B-2cosAcosBsinAsinB`
`=1-cos^2A-cos^2B-sin^2Asin^2B+cos^2Acos^2B`
`=sin^2A-cos^2B-sin^2Asin^2B+cos^2Acos^2B`
`=sin^2Acos^2B-cos^2Bsin^2A`
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