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` .

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

Using properties of determinants, show that ABC is isosceles if |(1,1,1),(1+cosA,1+cosB,1+cosC),(cos^2A+cosA,cos^2B+cosB,cos^2C+cosC)|=0

Using properties of determinants, show that ABC is isosceles if |(1,1,1),(1+cosA,1+cosB,1+cosC),(cos^2A+cosA,cos^2B+cosB,cos^2C+cosC)|=0

Using properties of determinants, show that triangle ABC is isosceles, if : |(1,1,1),(1+cosA, 1+cosB, 1+cosC),(sqrecosA+cosA, cossqrB+cosB, cossqurC+cosC)| =0

If A,B and C are angles of a triangle then the determinant |(-1,cosC,cosB),(cosC,-1,cosA),(cosB,cosA,-1)| is equal to

If A,B and C are angles of a triangle then the determinant |(-1,cosC,cosB),(cosC,-1,cosA),(cosB,cosA,-1)| is equal to

cos^2 A + cos^2 B +cos^2 C=1+2cosA cosB cosC .

If A,B and C are angles of a triangle, then the determinant: {:|(-1,cosC,cosB),(cosC,-1,cosA),(cosB,cosA,-1)| is equal to