Compute $\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}$ 
$$\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}=?$$

Here's what I have done so far:
$y=\frac{\pi}{2}-x$ The limit becomes:$$\lim_{y\to 0}\frac{(1-\cos y)(1-\cos^2y)\dots(1-\cos^ny)}{\sin^{2n}y}=\lim_{y\to 0}\frac{1-\cos y}{\sin^2y}\frac{1-\cos^2y}{\sin^2y}\dots\frac{1-\cos^ny}{\sin^2y}$$
Also $1-\cos y=2\sin^2 \frac{x}{2}$
How should i write $1-\cos^2y ,1-\cos^3y,\dots1-\cos^ny$  in order to get the final answer. 

Look into this
$\lim_{x\rightarrow 0} \frac{(1-\cos x)}{\sin^2 x} \cdot \frac{1-\cos^2 x}{\sin^2 x} \cdots \frac{1-\cos^n x}{\sin^2 x}$ the lim distribute over the fractions and each can be calculated using Lohpital rule
