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Question
Using properties of determinants, show that ΔABC is isosceles if:`|[1,1,1],[1+cosA,1+cosB,1+cosC],[cos^2A+cosA,cos^B+cosB,cos^2C+cosC]|=0`
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Solution
`|[1,1,1],[1+cosA,1+cosB,1+cosC],[cos^2A+cosA,cos^B+cosB,cos^2C+cosC]|=0`
Performing C2→C2−C1, C3→C3−C1
`|[1,0,0],[1+cosA,cosB-cosA,cosC-cosA],[cos^2A+cosA,(cosB-cosA)(cosA+cosB+1),(cosC−cosA)(cosC+cosA+1)]|=0`
`(cosB−cosA)(cosC−cosA)|[1,0,0],[1+cosA,1,1],[cos^2A+cosA,(cosA+cosB+1),(cosC+cosA+1)]|=0`
(cosB−cosA)(cosC−cosA)(cosC−cosB)=0
∴ cosB=cosA
⇒B=A or cosC=cosA
⇒C=A or cosC=cosB
⇒C=B
∴ △ABC is an isosceles triangle.
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