\begin{aligned}
& \text { We have to prove that, } \\
& \tan A+2 \tan 2 A+4 \tan 4 A+8 \cot 8 A=\cot A \\
& \text { i.e., to prove, } \\
& \cot A-\tan A-2 \tan 2 A-4 \tan 4 A-8 \cot 8 A=0 \\
& \cot \theta-\tan \theta=\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta} \\
& =\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos \theta \sin \theta} \\
& =\frac{2 \cos 2 \theta}{2 \sin \theta \cos \theta} \\
& =\frac{2 \cos 2 \theta}{\sin 2 \theta} \\
& \therefore \cot \theta-\tan \theta=2 \cot 2 \theta \ldots \text {...(i) } \\
& \text { L.H.S. }=\cot A-\tan A-2 \tan 2 A-4 \tan 4 A-8 \cot 8 A \\
& =2 \cot 2 A-2 \tan 2 A-4 \tan 4 A-8 \cot 8 A \ldots . .[\text { From (i)] } \\
& =2(\cot 2 A-\tan 2 A)-4 \tan 4 A-8 \cot 8 A \\
& =2 \times 2 \cot 2(2 A)-4 \tan 4 A-8 \cot 8 A \ldots \ldots[\text { [rom (i)] } \\
& =4(\cot 4 \mathrm{~A}-\tan 4 \mathrm{~A})-8 \cot 8 \mathrm{~A} \\
& =4 \times 2 \cot 2(4 \mathrm{~A})-8 \cot 8 \mathrm{~A} \ldots \ldots . \text { [From (i)] } \\
& =8 \cot 8 \mathrm{~A}-8 \cot 8 \mathrm{~A}=0 \\
& =\text { R.H.S. } \\
\end{aligned}