If sum of the series $\frac {\tan 1}{\cos 2}+\frac{\tan 2}{\cos 4} +\frac{\tan 3}{\cos 6}...\frac{\tan 1024}{\cos 2048}=\tan \lambda -\tan \mu$

If $\tan x,\tan2x$ are non-zero and finite, i.e., we need $2x\ne n90^\circ$ where $n$ is any integer

$$\dfrac{\tan x}{\cos2x}=\dfrac{\sin x}{\cos x\cos2x}=\dfrac{\sin(2x-x)}{\cos x\cos2x}=\tan2x-\tan x$$

So the problem should be $$\sum_{r=0}^{10}\dfrac{\tan(2^rx)}{\cos(2^{r+1}x)}$$