$\tan^2\large\frac{\theta}{2}=\frac{(1-\cos \theta)^2}{\sin^2 \theta}$
$\Rightarrow \large\frac{(1-\cos \theta)^2}{1-\cos^2\theta}$
$\Rightarrow \large\frac{1-\cos \theta}{1+\cos \theta}$
$\Rightarrow \large\frac{1-\Large\frac{\cos \alpha-\cos \beta}{1-\cos \alpha\cos\beta}}{1+\Large\frac{\cos \alpha-\cos \beta}{1-\cos \alpha\cos\beta}}$
$\Rightarrow \large\frac{1-\cos \alpha\cos\beta-\cos \alpha+\cos \beta}{1-\cos \alpha\cos \beta+\cos \alpha-\cos\beta}$
$\Rightarrow \large\frac{(1-\cos \alpha)+\cos\beta(1-\cos \alpha}{(1+\cos \alpha)-\cos \beta(1+\cos \alpha)}$
$\Rightarrow \large\frac{(1-\cos \alpha)(1+\cos \beta)}{(1+\cos \alpha)(1-\cos \beta)}$
$\Rightarrow \tan^2\large\frac{\alpha}{2}$$\cot^2\large\frac{\beta}{2}$
$\therefore \tan \large\frac{\theta}{2}=$$\pm \tan \large\frac{\alpha}{2}$$\tan\large\frac{ \beta}{2}$
Hence one of the values of $\tan\large\frac{\theta}{2}$ is $\tan \large\frac{\alpha}{2}$$\tan \large\frac{\beta}{2}$
Hence proved