Evaluating $\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan^2 \frac {a}{2^r} \tan \frac {a}{2^{r-1}}+\tan \frac {a}{2^{r}}}$ 
The question is to evaluate $$\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan^2 \frac {a}{2^r} \tan \frac {a}{2^{r-1}}+\tan \frac {a}{2^{r}}}$$

I could rewrite the denominator as $$\tan\frac {a}{2^r}\left(\tan \frac {a}{2^r}\tan \frac{a}{2^{r-1}}+1\right)$$which is same as $$\tan \frac{a}{2^r}-\tan\frac {a}{2^{r-1}}$$And so the product transforms to $$\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan \frac{a}{2^r}-\tan\frac {a}{2^{r-1}}}$$ I have no idea on how to proceed after this. Any help is appreciated. Thanks.

Lets start with:
$\tan \frac {a}{2^{r-1}} = \frac {2\tan \frac {a}{2^{r}}}{1-\tan^2 \frac {a}{2^{r}}}$
and that gets everything in terms of $\frac {a}{2^{r}}$
$u = \frac {a}{2^{r}}$
$\frac {\sin u}{\tan^2 u\frac {2 \tan u}{1-\tan^2 u}+\tan u}\\
\frac {\sin u (1-\tan^2 u)}{\tan u(1 +\tan^2 u)}\\
\frac {\sin u (1-\tan^2 u)}{\tan u(\sec^2 u)}\\
\cos^2 \cot u\sin u(1-\tan^2 u)\\
\cos u (\cos^2u-\sin^2 u)\\
\cos u (\cos 2u)$
$\prod_\limits{r=1}^{\infty} \cos \frac {a}{2^{r}} \cos \frac {a}{2^{r-1}} = \cos a\prod_\limits{r=1}^{\infty} \cos^2 \frac {a}{2^{r}}\\
\frac {\sin^2 a}{a^2} \cos {a}$
