Given that, $\sin (\theta+\alpha)=a$ and $\quad \sin (\theta+\beta)=b$
$\therefore \cos (\theta+\alpha)=\sqrt{1-a^2}$ and $\cos (\theta+\beta)=\sqrt{1-b^2}$
$\therefore \cos (\alpha-\beta)=\cos \{\theta+\alpha-(\theta+\beta)\}$
$=\cos (\theta+\beta) \cos (\theta+\alpha)+\sin (\theta+\alpha) \sin (\theta+\beta)$
$=\sqrt{1-a^2} \sqrt{1-b^2}+a \cdot b=a b+\sqrt{\left(1-a^2\right)\left(1-b^2\right)}$
$\Rightarrow \cos (\alpha-\beta)=a b+\sqrt{1-a^2-b^2+a^2 b^2}$
Now, $\cos 2(\alpha-\beta)-4 a b \cos (\alpha-\beta)$
$=2 \cos ^2(\alpha-\beta)-1-4 a b \cos (\alpha-\beta)$
$=2 \cos (\alpha-\beta)(\cos (\alpha-\beta)-2 a b)-1$
$=2\left(a b+\sqrt{1-a^2-b^2+a^2 b^2}\right)$
$\left(a b+\sqrt{1-a^2-b^2+a^2 b^2}-2 a b\right)-1$
$=2\left[\left(\sqrt{1-a^2-b^2+a^2 b^2}+a b\right)\right.$
$=2\left[1-a^2-b^2+a^2 b^2-a^2 b^2\right]-1$
$=2-2 a^2-2 b^2-1=1-2 a^2-2 b^2 \quad$ Hence proved.