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In a `triangleABC`, if `|[1,1,1][1+sinA,1+sinB,1+sinC],[sinA+sin^2A,sinB+sin^2B,sinC+sin^2C]|=0`, then prove that `triangleABC` is an isosceles triangle.

Answer» `|(1,1,1),(1+sinA , 1+ sinB, 1+sinC),(sinA +sin^2 A , sin B + sin^2 B , sin C + sin^2 C)|= 0`
`R_1 -> R_2 - R_3`
`|(1,1,1),(sinA , sinB, sinC),(sinA +sin^2 A , sin B + sin^2 B , sin C + sin^2 C)|= 0`
`R_3 -> R_3 -R_2`
`|(1,1,1),(sinA , sinB, sinC),(sin^2 A ,sin^2 B , sin^2 C)|= 0`
`C_2 -> C_2 - C_1 & C_3 -> C_3 - C_1`
`|(1,0,0),(sinA , sinB- sinA, sinC- sinB),(sin^2 A ,(sinB-sinA)(sinB+ sinA) , (sinC-sinB)(sin C + sinB))|= 0`
`(sinB-sinA)(sinC-sinA)|(11,0,0),(sinA,1,1),(sin^2 A , sinB+ sinA, sinC+sinA)|= 0`
`(sin B - sinA)(sin C-sinA)[1(sinC + sinA - sinB - sinA)] = 0`
`sinB = sinA or sinC = sinA or sinC= sinB`
`/_B =/_A ; /_C=/_A ; /_C= /_B`
so, `AC=BC , AB=BC,AB=AC`
`:./_ ABC` is isosceles . hence proved


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