Question

If $A,B$ and $C$ denote the angles of a triangle,
then $=\left|\begin{array}{ccc}-1& \cos C& \cos B\\ \cos C& -1& \cos A\\ \cos B& \cos A& -2\end{array}\right|$ independent of

• A. $A$
• B. $B$
• C. $C$
• D. None of these

Answer 1

Answer: (B)

$B$

Multiplying $C_1$ by $a$ and then applying
$C_1\to C_2+b{C}_2+C{C}_3$ we get
$\Delta =\frac{1}{a}\left|\begin{array}{ccc}-a+b\cos C+c\cos B& \cos C& \cos B\\ a\cos C-b-c\cos A& -1& \cos A\\ a\cos B+b\cos A-2c& \cos A& -2\end{array}\right|$
$\Rightarrow \Delta =\frac{1}{a}\left|\begin{array}{ccc}0& \cos C& \cos B\\ 0& -1& \cos A\\ -c& \cos A& -2\end{array}\right|$
$\Rightarrow \Delta =\frac{-c}{a}(\cos C\cos A+\cos B)$
$\Rightarrow \Delta =\frac{-c}{a}\sin C\sin A$, which is independent of $B$