<s>[INST] 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.
 [/INST] 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.
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