$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha =$

Given, $\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha$

Adding and Subtracting $\cot \alpha$ we get,

$\tan \alpha -\cot \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha +\cot \alpha$

$=-2\cot 2\alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha +\cot \alpha$  $\left[\because \tan \alpha -\mathrm{cot\alpha }=\frac{{\sin }^2\mathrm{\alpha }-{\cos }^2\mathrm{\alpha }}{\mathrm{sin\alpha cos\alpha }}=\frac{-2\cos 2\mathrm{\alpha }}{\sin 2\mathrm{\alpha }}=-2\cot 2\mathrm{\alpha }\right]$

$=-4\cot 4\alpha +4\tan 4\alpha +8\cot 8\alpha +\cot \alpha$

$=-8\cot 8\alpha +8\cot 8\alpha +\cot \alpha$

$=\cot \alpha$