Finding $\lim_{ n \to \infty }(1-\tan^2\frac{x}{2})(1-\tan^2\frac{x}{4})(1-\tan^2\frac{x}{8})...(1-\tan^2\frac{x}{2^m})=?$ Find the limit :
$$\lim_{ n \to \infty }(1-\tan^2\frac{x}{2})(1-\tan^2\frac{x}{4})(1-\tan^2\frac{x}{8})...(1-\tan^2\frac{x}{2^n})=?$$
My try :
$$1-\tan^2 y = \frac{2\tan y }{\tan(2y)}$$
$$\lim_{ n \to \infty }\left( \frac{2\tan\frac{x}{2} }{\tan(x)}\right)( \frac{2\tan\frac{x}{4} }{\tan(\frac{x}{2})})( \frac{2\tan\frac{x}{8} }{\tan(\frac{x}{4})})...( \frac{2\tan\frac{x}{2^n} }{\tan(\frac{x}{2^{n-1}})})=?$$
Now?

Note that the numbers in the deonominator and numerator cancel, leaving only $$\lim_{n \to \infty} 2^{n} \frac{\tan \frac{x}{2^n}}{\tan x}$$
In your original equation. However, as $\lim\limits_{n \to \infty} \frac{x}{2^n}=0$, we have that $$\lim_{n \to \infty} 2^n \tan \frac{x}{2^n}=\lim_{n \to \infty}x \times \dfrac{ \tan \frac{x}{2^n}}{\frac{x}{2^n}} =x$$ 
Using the fact that $\lim\limits_{a \to 0}\frac{\tan a}{a}=1$. So the limit becomes $$\lim_{n \to \infty} 2^n \times {\tan \frac{x}{2^n}} \times \frac{1}{\tan x}= \frac{x}{\tan x}=x \cot x$$
The answer is $x \cot x$. 
