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If A + B + C = 0, then prove that $\begin{array}{|ccc|} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array} =0$

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LHS given: $\begin{array}{|ccc|} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array}$

Expand along $R_1$

$\\\begin{array}{l} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{12}(-1)^{1+2}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{33} \end{array}\right| \\ +\mathrm{a}_{13}(-1)^{1+3}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{22} \\ \mathrm{a}_{31} & \mathrm{a}_{32} \end{array}\right| \\ =(1)\left|\begin{array}{cc} 1 & \cos \mathrm{A} \\ \cos \mathrm{A} & 1 \end{array}\right|-\cos \mathrm{C}\left|\begin{array}{cc} \cos \mathrm{C} & \cos \mathrm{A} \\ \cos \mathrm{B} & 1 \end{array}\right|+\cos \mathrm{B}\left|\begin{array}{cc} \cos \mathrm{C} & 1 \\ \cos \mathrm{B} & \cos \mathrm{A} \end{array}\right| \end{array}$

$\\\vspace{\baselineskip}= [1 \{ 1 -cos\textsuperscript{2}A \} -cos C \{ cos C -cos B cos A \} + cos B \{ cos C cos A -cos B \} ]\\ \\ \vspace{\baselineskip}= \{ 1 -cos\textsuperscript{2}A \} - \{ cos\textsuperscript{2}C -cos A cos B cos C \} + \{ cos A cos B cos C -cos\textsuperscript{2}B \} \\ \\ \vspace{\baselineskip}= \{ sin\textsuperscript{2}A \} -cos\textsuperscript{2}C + cos A cos B cos C + cos A cos B cos C -cos\textsuperscript{2}B\\ \\ \vspace{\baselineskip}[\because cos\textsuperscript{2}x + sin\textsuperscript{2}x = 1]\\ \\ \vspace{\baselineskip}= sin\textsuperscript{2}A -cos\textsuperscript{2}C -cos\textsuperscript{2}B + 2cos A cos B cos C\\ \\ \vspace{\baselineskip}= -(cos\textsuperscript{2}B -sin\textsuperscript{2}A)-cos\textsuperscript{2}C + 2cos A cos B cos C\\ \\$

$\\\vspace{\baselineskip}= -[cos(B + A)cos(B -A)] + cos C [2 cos A cos B -cos C]\\ \\ \vspace{\baselineskip}[\because cos\textsuperscript{2}B -sin\textsuperscript{2}A = cos(B + A)cos(B -A)]\\ \\ \vspace{\baselineskip}= -[cos(B + A)cos(B -A)] + cos C [cos(A + B) + cos(A -B) -cos C] \ldots (i)\\ \\ \vspace{\baselineskip}[\because 2cos A cos B = cos(A + B) + cos(A -B)]\\ (A+B+C) = 0, given\\ Which means, A + B = - C\\ \vspace{\baselineskip} = -[cos(- C) cos(B -A)] + cos C [cos(-C) + cos (A -B) -cos C]\\ \\ \vspace{\baselineskip}= -cos C cos(B -A) + cos C[cos C + cos(A -B) -cos C]\\ \\ \vspace{\baselineskip}[\because cos(-C) = cos C]\\ \\ \vspace{\baselineskip}Now, cos(A -B) = cos A cos B + sin A sin B\\ \\ \vspace{\baselineskip}= -cos C \{ cos B cos A + sin B sin A \} + cos C [cos A cos B + sin A sin B]\\ \\ \vspace{\baselineskip}= 0 = RHS\\$

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