Explanation

(b)
Given, angles of a triangle are $A,\ B\ and\ C$. We know that $A + B + C = \pi$, therefore
$A + B = \pi - C$
$\Rightarrow \cos{(A + B) = \cos{(\pi - C) = - \cos C}}$
$\Rightarrow \cos{A\cos B - \sin{A\sin{B = - \cos C}}}$
$\Rightarrow \cos{A\cos{B + \cos{C = \sin{A\sin B}}}}\$ …(i)
Let $\mathrm{\Delta} = \left| \begin{matrix}
- 1 & \cos C & \cos B \\
\cos C & - 1 & \cos A \\
\cos B & \cos A & - 1
\end{matrix} \right|$
$= - (1 - \cos^{2}{A) + \cos{C(\cos{C + \cos{A\cos{B) + \cos{B(\cos{B + \cos{A\cos{C)}}}}}}}}}$
$= - \sin^{2}{A + \cos{C(\sin{A\sin{B) + \cos{B(\sin{A\sin{C)}}}}}}}$ [from Eq.(i)]
$= - \sin^{2}{A + \sin{A(\sin{B\cos{C + \cos{B\sin{C)}}}}}}$
$= - \sin^{2}{A + \sin{A\ sin(B + C)}}$
$= - \sin^{2}{A + \sin^{2}{A = 0}}$ $\lbrack\because\sin{(B + C) = \sin{(\pi - A) = \sin{A\rbrack}}}$