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Using properties of determinant. Prove t...

Using properties of determinant. Prove that `|[sinA,cosA,sinA+cosB],[sinB,cosA,sinB+cosB],[sinC,cosA,sinC+cosB]|=0`

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Since `|(sinA,cosA,sinA+cosB),(sinB,cosA,sinB+cosB),(sinC,cosA,sinC+cosB)|=|(sinA,cosA,sinA),(sinB,cosA,sinB),(sinc,cosA,sincC)|+|(sincA,cosA,cosB),(sinB,cosA,cosB),(sinC,cosA,cosB)|`
`=0|(sinA,cosA,cosB),(sinB,cosA,cosB),(sinc,cosA,cosB)|`
[since in first determinant `C_(1)` and `C_(3)` are identicals]
`=cosA.cosB|(sinA,1,1),(sinB,1,1),(sinC,1,1)|`
[taking `cosA` common from `C_(2)` and `cosB` common from `C_(3)`]
`=0` [since `C_(2)` and `C_(3)` are identicals]
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