Question

If the value of determinant $\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1+\cos A& 1+\cos B& 1+\cos C\\ {\cos }^{2}A+\cos A& {\cos }^{2}B+\cos B& {\cos }^{2}C+\cos C\end{array}\right|$ is zero for $△ABC,$ whose angles are $A,B$ and $C.$ Then $△ABC$ can be

• A. Equilateral triangle
• B. Isosceles triangle
• C. Scalene triangle
• D. Right angled triangle, whose one angle is ${60}^{\circ }.$

Answer 1

Answer: (A), (B)

Isosceles triangle

Given : $\left|\begin{array}{ccc}1& 1& 1\\ 1+\cos A& 1+\cos B& 1+\cos C\\ {\cos }^{2}A+\cos A& {\cos }^{2}B+\cos B& {\cos }^{2}C+\cos C\end{array}\right|=0$

Applying $C_1\to C_1-C_3$ and $C_2\to C_2-C_3,$ we get
$\Rightarrow \left|\begin{array}{ccc}0& 0& 1\\ \cos A-\cos C& \cos B-\cos C& 1+\cos C\\ {\cos }^{2}A+\cos A-{\cos }^{2}C-\cos C& {\cos }^{2}B+\cos B-{\cos }^{2}C-\cos C& {\cos }^{2}C+\cos C\end{array}\right|=0$

Now, Taking $(\cos A-\cos C)$ common from $C_1$ and $(\cos B-\cos C)$ common from $C_2$, we get
$\Rightarrow (\cos A-\cos C)(\cos B-\cos C)\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& 1+\cos C\\ \cos A+\cos C+1& \cos B+\cos C+1& {\cos }^{2}C+\cos C\end{array}\right|=0$

Now expanding along $R_1,$ we get
$\Rightarrow (\cos A-\cos C)(\cos B-\cos C)(\cos B-\cos A)=0$
If $A=C$ or $B=C$ or $B=A$
Then $△ABC$ is an isosceles triangle.
If $A=B=C$
Then $△ABC$ is an equilateral triangle.