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Question

Show that ΔABC is an isosceles triangle, if the determinant
Δ=∣ ∣1111+cos A1+cos B1+cos Ccos2 A+cos Acos2 B+cos Bcos2 C+cos C∣ ∣=0

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Solution

We have, Δ=∣ ∣001cos Acos Ccos Bcos C1+cos Ccos2 A+cos Acos2 Acos2 Ccos Ccos2 B+cos Bcos2 Ccos Ccos2 C+cos C∣ ∣=0
[C1C1C3 and C2C2C3]
(cos Acos C).(cos Bcos C)∣ ∣001111+cos Ccos A+cos C+1cos B+cos C+1cos2 C+cos C∣ ∣=0
[taking (cos Acos C) common from C1 and (cos Bcos C) common from C2]
(cos Acos C).(cos Bcos C)[(cos B+cos C+1)(cos A+cos C+1)]=0(cos Acos C).(cos Bcos C)(cos B+cos C+1cos Acos C1)=0(cos Acos C).(cos Bcos C)(cos Bcos A)=0i.e.,cos A=cos C or cos B=cos C or cos B=cos A
A=C or B=C orB=A
Hence, ABC is an isosceles triangle.


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