Explanation

(a)
We know that,
$\alpha - \beta = (\theta - \beta) - (\theta - \alpha)$
$\therefore\cos{(\alpha - \beta) = \cos{(\theta - \beta)\cos{(\theta - \alpha) + \sin{(\theta - \beta)\sin{(\theta - \alpha)}}}}}$
$= ab + \sqrt{1 - a^{2}}\sqrt{1 - b^{2}}$
and$\ \sin{(\alpha - \beta) = \pm (a\sqrt{1 - b^{2}} - b\sqrt{1 - a^{2})}}$
$\Rightarrow \sin^{2}{(\alpha - \beta) = a^{2} + b^{2} - 2a^{2}b^{2} - 2ab\sqrt{1 - a^{2}}}\sqrt{1 - b^{2}}$
$\Rightarrow \sin^{2}{(\alpha - \beta) = a^{2} + b^{2} - 2a^{2}b^{2} - 2ab\left\lbrack \cos{(\alpha - \beta) - ab} \right\rbrack}$
$\therefore\sin^{2}{(\alpha - \beta) - a^{2} + b^{2} - 2ab\cos{(\alpha - \beta)}}$
$\Rightarrow \sin^{2}{(\alpha - \beta) + 2ab\cos{(\alpha - \beta) = a^{2} + b^{2}}}$