Write the argument as $\frac{2^{p-1}}{1+2^{2p-1}}=\frac{2^p-2^{p-1}}{1+2^p\cdot 2^{p-1}}$, which is the formula for ${\tan }^{-1}(2^p)-{\tan }^{-1}(2^{p-1})$.

So $\sum _{p=1}^{11}{\tan }^{-1}\left(\frac{2^{p-1}}{1+2^{2p-1}}\right)={\tan }^{-1}(2^{11})-{\tan }^{-1}(2^0)={\tan }^{-1}(2048)-\frac{\pi }{4}$.

Therefore $\alpha =\frac{\pi }{4}+{\tan }^{-1}(2048)-\frac{\pi }{4}={\tan }^{-1}(2048)$, and $\tan \alpha =2048$.