Is there easier way out?

We have,

$\displaystyle tan\left( \frac{\theta+\alpha}{2}\right)tan\left(\frac{\theta-\alpha}{2}\right)=f(\beta)$

Simplifying, we get:

$\displaystyle f(\beta) = \frac{sin\left( \frac{\theta+\alpha}{2}\right)sin\left(\frac{\theta-\alpha}{2}\right)}{cos\left(\frac{\theta+\alpha}{2}\right)cos\left(\frac{\theta-\alpha}{2}\right)}$

Multipying and dividing by $-2$ , we get:

$\displaystyle f(\beta) = \frac{-2sin\left(\frac{\theta+\alpha}{2}\right)sin\left(\frac{\theta-\alpha}{2}\right)}{-2cos\left(\frac{\theta+\alpha}{2}\right)cos\left(\frac{\theta+\alpha}{2}\right)}$

Which is equivalent to,

$\displaystyle f(\beta) = \frac{cos\theta - cos\alpha}{-(cos\theta + cos\alpha)}$

Using $cos(\theta)=cos(\alpha)cos(\beta)$ , we get :

$\displaystyle f(\beta) = \frac{cos\alpha(cos\beta -1)}{-cos\alpha(cos\beta +1)} = \frac{1-cos\beta}{1+cos\beta}$

Hence,

$\displaystyle f\left(\frac{\pi}{6}\right) = \frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}} = 7 - 4\sqrt{3} \approx \boxed{0.0718}$

Let $t_0 = \tan \frac{\theta}{2}$ , $t_1 = \tan \frac{\alpha}{2}$ and $t_2 = \tan \frac{\beta}{2}$ . Then, we have:

$\begin{aligned} f(\beta) & = \tan \left( \frac{\theta + \alpha}{2} \right) \tan \left( \frac{\theta - \alpha}{2} \right) \\ & = \frac{t_0+t_1}{1-t_0t_1} \dot{} \frac{t_0-t_1}{1+t_0t_1} \\ & = \frac{t_0^2-t_1^2}{1-t_0^2t_1^2} \end{aligned}$

$\begin{aligned} \cos \theta & = \cos{\alpha} \cos{\beta} \\ \frac{1+t_0^2}{1-t_0^2} & = \frac{1+t_1^2}{1-t_1^2} \dot{} \frac{1+t_2^2}{1-t_2^2} \\ \frac{1+\color{#3D99F6}{t_2^2}}{1-\color{#3D99F6}{t_2^2}} & = \frac{1+t_0^2}{1-t_0^2} \dot{} \frac{1-t_1^2}{1+t_1^2} \\ & = \frac{1+t_0^2-t_1^2-t_0^2t_1^2}{1-t_0^2+t_1^2-t_0^2t_1^2} \quad \quad \color{#3D99F6}{\text{Divide nominator and denominator by } 1 - t_0^2t_1^2} \\ & = \frac{1+\color{#3D99F6}{f(\beta)}}{1-\color{#3D99F6}{f(\beta)}} \end{aligned}$

$\begin{aligned} \Rightarrow f(\beta) & = t_2^2 = \tan^2 \frac{\beta}{2} \\ \Rightarrow f \left(\frac{\pi}{6} \right) & = \tan^2 \frac{\pi}{12} = \boxed{0.0718} \end{aligned}$