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.
 A: The product can be rewritten as
$$\prod_{r=1}^{\infty} \frac{\cos{\left ( 2^{-r} a \right )}}{1+\tan{\left ( 2^{-r} a \right )} \tan{\left ( 2^{-r+1} a \right )}} $$
Use the tangent half-angle formula to simplify the denominator:
$$\begin{align}1+\tan{\left ( 2^{-r} a \right )} \tan{\left ( 2^{-r+1} a \right )} &= 1+\sqrt{\frac{1-\cos{\left ( 2^{-r} a \right )}}{1+\cos{\left ( 2^{-r} a \right )}}} \frac{\sqrt{1-\cos^2{\left ( 2^{-r} a \right )}}}{\cos{\left ( 2^{-r} a \right )}}\\ &= \frac1{\cos{\left ( 2^{-r} a \right )}} \end{align}$$
The product is then equal to
$$\left [ \prod_{r=1}^{\infty} \cos{\left ( 2^{-r} a \right )} \right ]^2 $$
Note that
$$ \prod_{r=1}^{\infty} \cos{\left ( 2^{-r} a \right )} = \frac{\sin{a}}{a}$$
Therefore, the product is equal to
$$\frac{\sin^2{a}}{a^2} $$
ADDENDUM
The product in the square brackets may be evaluated with ease using the sine double angle forumla.  The product of the cosines is equal to
$$\lim_{n \to \infty} \frac{\sin{a}}{2 \sin{(a/2)}} \frac{\sin{(a/2)}}{2 \sin{(a/2)}} \cdots \frac{\sin{(a/2^{n-1})}}{2 \sin{(a/2^n)}} = \lim_{n \to \infty} \frac{\sin{a}}{2^n \sin{(a/2^n)}}$$
The result follows.
A: 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 $\frac {a}{2^{r}}$
$u = \frac {a}{2^{r}}$
$\frac {\sin u}{\tan^2 u\frac {2 \tan u}{1-\tan^2 u}+\tan u}\\
\frac {\sin u (1-\tan^2 u)}{\tan u(1 +\tan^2 u)}\\
\frac {\sin u (1-\tan^2 u)}{\tan u(\sec^2 u)}\\
\cos^2 \cot u\sin u(1-\tan^2 u)\\
\cos u (\cos^2u-\sin^2 u)\\
\cos u (\cos 2u)$
$\prod_\limits{r=1}^{\infty} \cos \frac {a}{2^{r}} \cos \frac {a}{2^{r-1}} = \cos a\prod_\limits{r=1}^{\infty} \cos^2 \frac {a}{2^{r}}\\
\frac {\sin^2 a}{a^2} \cos {a}$
