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

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 $\Delta =\left|\begin{array}{ccc}-1& \cos C& \cos B\\ \cos C& -1& \cos A\\ \cos B& \cos A& -1\end{array}\right|$
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 ${\cos }^{2}A+{\sin }^{2}A=1]$
$=-{\sin }^{2}A+\sin A(\sin B\cos C+\cos B\sin C)$
$=-{\sin }^{2}A+\sin A\sin (B+C)$
$[\because \sin (x+y)=\sin x\cos y+\cos x\sin y]$
$=-{\sin }^{2}A+{\sin }^{2}A=0\text{ [using Eq. (iii)] }$