Show that `DeltaABC` is an isosceles triangle, if the determinant
`Delta=|(1,1,1),(1+cosA,1+cosB,1+cosC),(cos^(2)A+cosA,cos^(2)B+cosB,cos^(2)C+cosC)|=0` .

Answer

N/a

Explanation

We have `Delta=|(1,1,1),(1+cosA,1+cosB,1+cosC),(cos^(2)A+cosA,cos^(2)B+cosB,cos^(2)C+cosC)|=0`
`Delta=|[0,0,1],[cosA-cosC,cosB-cosC,1+cosC],[cos^(2)A+cosA-cos^(2)C-cosC,cos^(2)B+cosB-cos^(2)C-cosC,cos^(2)C+cosC]|=0`
`[ :' C_(1)toC_(1)-C_(3)` and `C_(2)toC_(2)-C_(3)]`
`implies (cosA-cosC).(cosB-cosC)`
`implies (cosA-cosC).(cosB-cosC)`
`|(0,0,1),(1,1,1+cosC),(cosA+cosC+1,cosB+cosC+1,cos^(2)C+cosC)|=0`
[taking `(cosA-cosC)` common from `C_(1)` and `(cosB-cosC)` common from `C_(2)`]
`implies (cosA-cosC).(cosB-cosC)[(cosB+cosC+1)-(cosA+cosC+1)]=0`
`= (cosA-cosC).(cosB-cosC)(cosB+cosC+1-cosA-cosC-1)=0`
`implies(cosA-cosC).(cosB-cosC)(cosB-cosA)=0`
i.e., `cosA =cosC` or `cosB=cosC` or `cosB=cosA`
`implies A=C` or `B=C` or `B=A`
Hence `ABC` is an isosceles triangle.