The value of the following product is? 
Evaluate the following product:
$$\newcommand{\T}[1]{\frac{\sin\frac{\theta}{#1}}{\tan^2\frac{\theta}{#1}\tan\frac{2\theta}{#1} + \tan\frac{\theta}{#1}}} \\
P(\theta) = \T{2} \times \T{2^2} \times \T{2^3} \times .... \infty$$
For $\theta = \frac \pi 4$

Simplified, $P(\theta)$ is
$$P(\theta) = \lim_{n \to \infty}\prod_{r=1}^n T(\theta,r)= \lim_{n \to \infty}\prod_{r=1}^n\T{2^r}$$
The denominator can be simplified as follows:
$$D = \tan\frac{\theta}{2^r}\left( \tan\frac{\theta}{2^r}\tan\frac{\theta}{2^{r-1}} + 1\right) \\
= \tan\frac{\theta}{2^{r-1}} - \tan\frac{\theta}{2^{r}}$$
After this, $P(\theta)$ becomes
$$P(\theta) = \lim_{n \to \infty}\prod_{r=1}^n \frac{\sin\frac{\theta}{2^r}}{\tan\frac{\theta}{2^{r-1}}- \tan\frac{\theta}{2^r}}$$
One more detail I found out is that $\lim_{n \to \infty} T(\theta,n) = 1$, but I couldn't proceed further from here. Any hints/solutions are appreciated.
EDIT: After the hints in the comments, $T(\theta, r)$ resolves to $\cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r}$ as follows (assuming $\frac \theta {2^r} = t$)
$$\begin{gather}
T(\theta, n) = \frac{\sin t}{\tan^2t\tan 2t + \tan t} \\
= \frac{\cos t}{\tan t \tan 2t + 1} \\
= \frac{\cos t(1-\tan^2t)}{1+\tan^2t} \\
= \cos t \cos 2t \\
= \cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r}
\end{gather}$$
Now,
$$P(\theta) = \lim_{n \to \infty} \frac{ \left( \cos\theta\cos\frac\theta2... \cos \frac{\theta}{2^n} \right)^2 }{\cos\theta} = \frac{\sin^2\theta}{2^{2n}\sin^2 \frac \theta {2^n}\cos \theta} = \frac{\sin^2 \theta}{\theta^2 \cos \theta}$$
Therefore,
$$\boxed{P(\pi/4) = \frac{8\sqrt2}{\pi^2}}$$
However, the answer mentioned in the textbook is $\frac{2}{\pi}$. Where am I going wrong? (I think there's a silly mistake somewhere here; just not able to find it :(
 A: Carrying on from the answer:
$$\begin{gather}
T(\theta, n) = \frac{\sin t}{\tan^2t\tan 2t + \tan t} \\
= \frac{\cos t}{\tan t \tan 2t + 1} \\
= \frac{\cos t(1-\tan^2t)}{1+\tan^2t} \\
= \cos t \cos 2t \\
= \cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r}
\end{gather}$$
Now,
$$P(\theta) = \cos\theta \cos \frac \theta 2 \cdot \cos\frac\theta2 \cos\frac\theta{2^2}\cdot ... = \left( \cos\theta \cos\frac\theta2...\right) \left( \cos\frac\theta2\cos\frac\theta{2^2}...\right)$$
Let
$$\begin{gather}
S = \lim_{n \to \infty}\cos\frac\theta2...\cos\frac\theta{2^n}\\
S\sin\frac\theta{2^n} = \lim_{n \to \infty} \frac{\sin\theta}{2^n} \\
S = \lim_{n \to \infty} \frac{\sin\theta}{2^n \sin\frac{\theta}{2^n}} = \lim_{t \to 0} \frac{t\sin\theta}{\sin(\theta t)}\\
S = \frac{\sin\theta}{\theta}
\end{gather}$$
Therefore, $P$ reduces to
$$P(\theta) = \frac{\sin2\theta\sin\theta}{2\theta^2}$$
And the value of $P(\pi/4)$ would be
$$P(\pi/4) = \frac{16}{2\sqrt2\pi^2}\\
\boxed{P(\pi/4) = \frac{4\sqrt2}{\pi^2}}$$
