How to prove that $\tan^2(\frac\theta2)= \tan^2(\frac\alpha2)\tan^2(\frac\beta2)$? I'm unable to solve this question:
$\cos(\theta)=\dfrac{\cos(\alpha)+\cos(\beta)}{1+\cos(\alpha) \cos(\beta)}$
Prove: $\tan^2\left(\frac\theta2\right)= \tan^2\left(\frac\alpha2\right)\tan^2\left(\frac\beta2\right)$
I have tried  the following:


*

*Using the indentity: $\cos(\theta)=\dfrac{1-\tan^2\left(\frac\theta2\right)}{1+\tan^2\left(\frac\theta2\right)}$

*Diving by $\cos (\alpha)  \cos (\beta)$

*Creating a triangle, to find $\tan(\theta)= \sin \alpha  \sin \beta$
Every time I got a huge complex equation with roots. Any help would be appreciated. 

$$ \tan ^2 {\frac {\theta}{2}} = \frac{\sin^2 {\frac {\theta}{2}} }{\cos^2 {\frac {\theta}{2}}} = \frac{ \left({\dfrac{1 - \cos \theta}{2}}\right)}{\left({\dfrac{1 + \cos \theta}{2}}\right)} \;\; \text{(since $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta= 1 - 2 \sin^2 \theta = 2 \cos^2\theta - 1$ )}$$
Now substituting for $\cos \theta$ and factorising we have, 
$$ = \frac{1 + \cos \alpha \cos \beta -  \cos \alpha - \cos \beta }{ 1 + \cos \alpha \cos \beta +  \cos \alpha + \cos \beta} = \frac{(1 - \cos \alpha)(1 - \cos \beta)}{(1 + \cos \alpha)(1 + \cos \beta)} $$
Now use the same formulas for $cos 2 \theta $ mentioned above strategically to negate the $1$ present in all the factors and to introduce $\dfrac {\alpha}{2} $. Then, 
$$ = \frac{[1 - (1 - 2 \sin^2 \frac {\alpha}{2} ) ][1 - ( 1 - 2 \sin^2 \frac {\beta}{2})]}{[1 + (2 \cos^2 \frac {\alpha}{2} - 1)][1 + (2 \cos ^2 \frac {\beta}{2}  - 1)]} $$
Remember $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$.
Now Simplify!!
