Explanation

(a)
Given,
$\left| \begin{matrix}
1 & cos(\alpha - \beta) & cos\ \alpha \\
cos(\alpha - \beta) & 1 & \cos\beta \\
\cos\alpha & cos\ \beta & 1
\end{matrix} \right|$ is symmetric determinant.
$\therefore$ Its value is
$1 + 2\cos(\alpha - \beta)\cos{\alpha\cos\beta}$
$- \cos^{2}{\alpha - \cos^{2}{\beta - \cos^{2}{(\alpha - \beta)}}}$
$= 1 - \cos^{2}{\alpha - \cos^{2}{\beta - cos(\alpha - \beta)}}$
$\lbrack\cos(\alpha - \beta) - 2\cos{\alpha\cos{\beta\rbrack}}$
$= 1 - cos^{2}\ \alpha - \cos^{2}{\beta - cos(\alpha - \beta)}$
$\left\lbrack \cos(\alpha - \beta) - \cos(\alpha + \beta) - \cos(\alpha - \beta) \right\rbrack$
$= 1 - \cos^{2}{\alpha - \cos^{2}{\beta + \cos(\alpha - \beta)\cos{(\alpha + \beta)}}}$
$= 1 - \cos^{2}{\alpha - \cos^{2}{\beta + \cos^{2}{\alpha\cos^{2}{\beta - \sin^{2}{\alpha\sin^{2}\beta}}}}}$
$= 1 - \cos^{2}{\alpha - \cos^{2}{\beta(1 - \cos^{2}{\alpha) - \sin^{2}{\alpha\sin^{2}\beta}}}}$
$= (1 - \cos^{2}{\alpha)(1 - \cos^{2}{\beta) - \sin^{2}{\alpha\sin^{2}\beta}}}$
$= \sin^{2}{\alpha\sin^{2}{\beta - \sin^{2}{\alpha\sin^{2}{\beta = 0}}}}$