सिद्ध कीजिए -
`|(sin^2A,sinA,cos^2A),(sin^2B,sinB,cos^2B),(sin^2C,sinC,cos^2C)|=-(sinA-sinB)(sinB-sinC)(sinC-sinA)`

Explanation

L.H.S.`=|(sin^2A,sinA,cos^2A),(sin^2B,sinB,cos^2B),(sin^2C,sinC,cos^2C)|`
`=|(sin^2A,sinA,1-sin^2A),(sin^2B,sinB,1-sin^2B),(sin^2C,sinC,1-sin^2C)|`
`=|(sin^2A,sinA,1),(sin^2B,sinB,1),(sin^2C,sinC,1)|`
`=|(sin^2A-sin^2C,sinA-sinC,0),(sin^2B-sin^2C,sinB-sinC,0),(sin^2C,sinC,1)|`
(संक्रियाओं `R_1toR_1-R_3` और `R_2toR_2-R_3` से )
`=|(sinA+sinC,1,0),(sinB+sinC,1,0),(sin^2C,sinC,1)|`
[`R_1` और `R_2` से क्रमशः और `(sinB-sinC)` लेने पर ]
`=(sinA-sinC)(sinB-sinC)`
`=|(sinA+sinC,,0),(sinB+sinC,,1)|`
(`C_3` के अनुदिश प्रसार करने पर )
`=(sinA-sinC)(sinB-sinC)(sinA+sinC-sinB-sinC)`
`=-(sinA-sinB)(sinB-sinC)(sinC-sinA)`
R.H.S.