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
We have, Δ=∣∣
∣∣001cos A−cos Ccos B−cos C1+cos Ccos2 A+cos A−cos2 A−cos2 C−cos Ccos2 B+cos B−cos2 C−cos Ccos2 C+cos C∣∣
∣∣=0
[∴C1→C1−C3 and C2→C2−C3]
⇒(cos A−cos C).(cos B−cos C)∣∣
∣∣001111+cos Ccos A+cos C+1cos B+cos C+1cos2 C+cos C∣∣
∣∣=0
[taking (cos A−cos C) common from C1 and (cos B−cos C) common from C2]
⇒(cos A−cos C).(cos B−cos C)[(cos B+cos C+1)−(cos A+cos C+1)]=0⇒(cos A−cos C).(cos B−cos C)(cos B+cos C+1−cos A−cos C−1)=0⇒(cos A−cos C).(cos B−cos C)(cos B−cos 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.