Question

Prove that $\left|\begin{array}{ccc}-1+\cos B& \cos C+\cos B& \cos B\\ \cos C+\cos A& -1+\cos A& \cos A\\ -1+\cos B& -1+\cos A& -1\end{array}\right|$

Answer 1

We have,

$D=\left|\begin{array}{ccc}-1+\cos B& \cos C+\cos B& \cos B\\ \cos C+\cos A& -1+\cos A& \cos A\\ -1+\cos B& -1+\cos A& -1\end{array}\right|$

$D=\left|\begin{array}{ccc}\cos B-1& \cos C+\cos B& \cos B\\ \cos C+\cos A& \cos A-1& \cos A\\ \cos B-1& \cos A-1& -1\end{array}\right|$

Applying $R_1\to R_1+R_2$

$D=\left|\begin{array}{ccc}(\cos A+\cos B+\cos C-1)& (\cos A+\cos B+\cos C-1)& \cos A+\cos B\\ \cos C+\cos A& -1+\cos A& \cos A\\ -1+\cos B& -1+\cos A& -1\end{array}\right|$

Applying $C_1\to C_1-C_2$ and $C_2\to C_2-C_1$

$D=\left|\begin{array}{ccc}0& 0& \cos A+\cos B\\ 1+\cos C& -(1+\cos C)& \cos A\\ \cos B-\cos A& \cos A-\cos B& -1\end{array}\right|$

$D=\left|\begin{array}{ccc}0& 0& \cos A+\cos B\\ 1+\cos C& -(1+\cos C)& \cos A\\ -(\cos A-\cos B)& \cos A-\cos B& -1\end{array}\right|$

On expanding w.r.t $R_1$, we get

$D=(\cos A+\cos B)\left[\right(1+\cos C\left)\right(\cos A-\cos B)-(1+\cos C\left)\right(\cos A-\cos B\left)\right]$

$D=(\cos A+\cos B)\times 0$

$D=0$

Hence, this is the answer.