Explanation

(a)
$\alpha - \beta = (\theta - \beta) - (\theta - \alpha)$
$\therefore\ \cos{(\alpha - \beta) = \cos{(\theta - \beta)\cos{(\theta - \alpha) + \sin{(\theta - \beta)}\sin{(\theta - \alpha)}}}}$
And $\sin{(\alpha - \beta) = {sin\ }{(\theta - \beta){cos\ }{(\theta - \alpha) - {sin\ }{(\theta - \alpha)}{cos\ }{(\theta - \beta)}}}}$
$\Rightarrow$ ${cos\ }{(\alpha - \beta)} = b.a + \sqrt{1 - a^{2}}\sqrt{1 - b^{2}}$
And $\sin{(\alpha - \beta) = \left( a\sqrt{1 - b^{2}} \right) - \left( b\sqrt{1 - b^{2}} \right)}$
Now, $\sin{{\ ^{2}(\alpha - \beta)}^{\ } = \left( a\sqrt{1 - b^{2}} \right)^{2} + \left( b\sqrt{1 - b^{2}} \right)^{2} - 2ab\sqrt{1 - a^{2}}\sqrt{1 - b^{2}}}$
$\Rightarrow \sin^{2}{(\alpha - \beta) = a^{2}\left( 1 - b^{2} \right) + b^{2}\left( 1 - a^{2} \right) - 2ab\left\{ \cos{(\alpha - \beta) - ab} \right\}}$
$\Rightarrow \sin^{2}{(\alpha - \beta) + 2ab\cos{(\alpha - \beta) = a^{2} - a^{2}b^{2} + b^{2} - b^{2}a^{2} + 2a^{2}b^{2}}}$
$\Rightarrow \ \sin^{2}{(\alpha - \beta) + 2ab\cos{(\alpha - \beta) = a^{2} + b^{2}}}$