$\cos \theta=\cos \alpha.\cos\beta$
$\therefore \cos \beta=\large\frac{\cos \theta}{\cos \alpha}$-----(1)
RHS
$\tan^2\large\frac{\beta}{2}=\frac{(1-\cos \beta)^2}{\sin ^2\beta}$
$\Rightarrow \large\frac{(1-\cos \beta)^2}{(1-\cos^2\beta)}=\frac{(1-\cos \beta)^2}{1-\cos\beta)(1+\cos \beta)}$
$\Rightarrow \large\frac{1-\cos \beta}{1+\cos \beta}=\frac{1-\Large\frac{\cos \theta}{\cos \alpha}}{1+\Large\frac{\cos \theta}{\cos \alpha}}$
$\Rightarrow \large\frac{\cos \alpha-\cos \theta}{\cos \alpha+\cos \theta}$
$\Rightarrow \large\frac{2\sin \large\frac{\alpha+\theta}{2}\sin \Large\frac{\theta-\alpha}{2}}{2\cos\Large\frac{\theta+\alpha}{2}\cos\Large\frac{\theta-\alpha}{2}}$
$\Rightarrow \tan \large\frac{\theta+\alpha}{2}.\frac{\theta-\alpha}{2}$
Hence proved