Question

If $A, B$ and $C$ are angles of a triangle, then the determinant $\begin{vmatrix}-1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1\end{vmatrix}$ is equal to

• A. 0
• B. $-1$
• C. 1
• D. None of these

Answer 1

Answer: (A)

Given $A, B$ and $C$ are the angles of a triangle.
$\therefore A+B+C =\pi $
$\Rightarrow A+B =\pi-C$
Now, $\cos (A+B)=\cos (\pi-C)=-\cos C$
$\Rightarrow \cos A \cos B-\sin A \sin B=-\cos C$
$\Rightarrow \cos A \cos B+\cos C=\sin A \sin B...$(i)
Similarly, $\cos A \cos C+\cos B=\sin A \sin C...$(ii)
and $ \sin (B+C)=\sin (\pi-A)=\sin A...$(iii)
Now, let $ \Delta=\begin{vmatrix}-1 & \cos C & \cos B \\\cos C & -1 & \cos A \\\cos B & \cos A & -1\end{vmatrix}$
By expanding along $R_1$, we get
$\Delta =-1\left(1-\cos ^2 A\right)+\cos C(\cos C+\cos A \cos B) $
$ +\cos B(\cos B+\cos A \cos C)$
$= -\sin ^2 A+\cos C(\sin A \sin B)+\cos B(\sin A \sin C)$
[using Eqs. (i) and (ii); and $\left. \cos ^2 A+\sin ^2 A=1\right] $
$= -\sin ^2 A+\sin A(\sin B \cos C+\cos B \sin C)$
$= -\sin ^2 A+\sin A \sin (B+C) $
$ [\because \sin (x+y)=\sin x \cos y+\cos x \sin y] $
$= -\sin ^2 A+\sin ^2 A=0 \text { [using Eq. (iii)] }$