Question

If $A+B+C=0,$ then value of $\left|\begin{array}{ccc}1& \cos C& \cos B\\ \cos C& 1& \cos A\\ \cos B& \cos A& 1\end{array}\right|$ is

Answer 1

Given, $\left|\begin{array}{ccc}1& \cos C& \cos B\\ \cos C& 1& \cos A\\ \cos B& \cos A& 1\end{array}\right|$
$=1(1-{\cos }^{2}A)-\cos C(\cos C-\cos A\cdot \cos B)+\cos B(\cos C\cdot \cos A-\cos B)={\sin }^{2}A-{\cos }^{2}C+\cos A\cdot \cos B\cdot \cos C+\cos A\cdot \cos B\cdot \cos C-{\cos }^{2}B={\sin }^{2}A-{\cos }^{2}B+2\cos A\cdot \cos B\cdot \cos C-{\cos }^{2}C=-\cos (A+B)\cdot \cos (A-B)+2\cos A\cdot \cos B\cdot \cos C-{\cos }^{2}C[\because {\cos }^{2}B-{\sin }^{2}A=\cos (A+B)\cdot \cos (A-B\left)\right]=-\cos (-C)\cdot \cos (A-B)+\cos C(2\cos A\cdot \cos B-\cos C)=-\cos C(\cos A\cdot \cos B+\sin A\cdot \sin B-2\cos A\cdot \cos B+\cos C)=\cos C(\cos A\cdot \cos B-\sin A\cdot \sin B-\cos C)=\cos C\left[\cos \right(A+B)-\cos C]$ $=\cos C(\cos C-\cos C)\{\because \cos C=\cos (A+B\left)\right\}=0$