Question

If A, B and C are angles of a triangle, then the determinant $\left|\begin{array}{ccc}-1& \cos C& \cos B\\ \cos C& -1& \cos A\\ \cos B& \cos A& -1\end{array}\right|$ is equal to
(a) 0
(b) –1
(c) 1
(d) none of these

Answer 1

Given: A, B and C are angles of a triangle
Therefore, $A+B+C=\mathrm{\pi }...\left(1\right)$


$$\begin{gathered} \left|\begin{array}{ccc}-1& \cos C& \cos B\\ \cos C& -1& \cos A\\ \cos B& \cos A& -1\end{array}\right| \\ \mathrm{Expanding}\mathrm{through}{R}_1 \\ =\left\{-1\left(1-{\cos }^{2}A\right)-\cos C\left(-\cos C-\cos A\cos B\right)+\cos B\left(\cos A\cos C+\cos B\right)\right\} \\ =-1+{\cos }^{2}A+{\cos }^{2}C+\cos A\cos B\cos C+\cos A\cos B\cos C+{\cos }^{2}B \\ =-1+2\cos A\cos B\cos C+{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C \\ =-1+2\cos A\cos B\cos C+\frac{1+\cos 2A}{2}+\frac{1+\cos 2B}{2}+{\cos }^{2}C\left(\because 2{\cos }^2\theta =1+\cos 2\theta \right) \\ =-1+2\cos A\cos B\cos C+\frac{1+\cos 2A+1+\cos 2B}{2}+{\cos }^{2}C \\ =-1+2\cos A\cos B\cos C+\frac{2+2\cos \left({\displaystyle \frac{2A+2B}{2}}\right)\cos \left({\displaystyle \frac{2A-2B}{2}}\right)}{2}+{\cos }^{2}C\left(\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\right) \\ =-1+2\cos A\cos B\cos C+\frac{2+2\cos \left({\displaystyle A+B}\right)\cos \left({\displaystyle A-B}\right)}{2}+{\cos }^{2}C \\ =-1+2\cos A\cos B\cos C+\frac{2+2\cos \left({\displaystyle \mathrm{\pi }-\mathrm{C}}\right)\cos \left({\displaystyle A-B}\right)}{2}+{\cos }^{2}C\left(\because A+B+C=\mathrm{\pi }\right) \\ =-1+2\cos A\cos B\cos C+\frac{2-2\cos C\cos \left({\displaystyle A-B}\right)+2{\cos }^{2}C}{2} \\ =-1+2\cos A\cos B\cos C+1-\cos C\cos \left(A-B\right)+{\cos }^{2}C \\ =-1+2\cos A\cos B\cos C+1-\cos C\left(\cos \left(A-B\right)-\cos C\right) \\ =-1+2\cos A\cos B\cos C+1-\cos C\left(-2\sin \left(\frac{A-B+C}{2}\right)\sin \left(\frac{A-B-C}{2}\right)\right)\left(\because \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\right) \\ =-1+2\cos A\cos B\cos C+1-\cos C\left(-2\sin \left(\frac{A+C-B}{2}\right)\sin \left(\frac{A-\left(B+C\right)}{2}\right)\right) \\ =-1+2\cos A\cos B\cos C+1-\cos C\left(-2\sin \left(\frac{\mathrm{\pi }-B-B}{2}\right)\sin \left(\frac{A-\left(\mathrm{\pi }-A\right)}{2}\right)\right)\left(\because A+B+C=\mathrm{\pi }\right) \\ =-1+2\cos A\cos B\cos C+1+2\cos C\sin \left(\frac{\mathrm{\pi }}{2}-B\right)\sin \left(A-\frac{\mathrm{\pi }}{2}\right) \\ =-1+2\cos A\cos B\cos C+1+2\cos C\cos B\sin \left(A-\frac{\mathrm{\pi }}{2}\right) \\ =-1+2\cos A\cos B\cos C+1-2\cos C\cos B\sin \left(\frac{\mathrm{\pi }}{2}-A\right) \\ =-1+2\cos A\cos B\cos C+1-2\cos C\cos B\cos A \\ =0 \end{gathered}$$


Hence, the correct option is (a).