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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​ - Mathematics

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 C2C2C1, C3C3C1

`|[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`

(cosBcosA)(cosCcosA)(cosCcosB)=0 

 cosB=cosA

B=or cosC=cosA

C=or cosC=cosB

C=B

 ABC is an isosceles triangle.

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