Solve $\tanα+2\tan2α+4\tan4α+8\tan8α+16\tanα=\cotα$ for $\alpha$ My knowledge of trigonometry are still insufficient to resolve this problem. Any help would be greatly appreciated.
Solving for $\alpha$:
$$\tanα+2\tan2α+4\tan4α+8\tan8α+16\tanα=\cotα$$
 A: As $\cos2A=\cos^2A-\sin^2A, \sin2A+2\sin A\cos A,$
$$\cot A-\tan A=\frac{\cos^2A-\sin^2A}{\cos A\sin A}=2\cot 2A$$
$$2\tan2\alpha+4\tan4\alpha+8\tan8\alpha+16\tan\alpha=\cot\alpha-\tan\alpha=2\cot2\alpha$$
$$\implies 4\tan4\alpha+8\tan8\alpha+16\tan\alpha=2(\cot2\alpha-\tan2\alpha)=2\cdot2\cot4\alpha$$
$$\implies 8\tan8\alpha+16\tan\alpha=4(\cdot2\cot4\alpha-\tan4\alpha)=4\cdot(2\cot8\alpha)$$
$$\implies 16\tan\alpha=8\cdot(\cot8\alpha-\tan8\alpha)=8\cdot2\cot16\alpha$$
$$\implies \cot16\alpha=\tan\alpha=\cot\left(\frac\pi2-\alpha\right)$$
$$\implies 16\alpha=n\pi+\frac\pi2-\alpha$$ where $n$ is any integer
$$\implies \alpha=\frac{(2n+1)\pi}{2\cdot17}$$ 
A: While there may be a more slick way to solve a problem like this, you can repeatedly apply $$\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$$ until you have a rational equation in $\tan(\alpha)$. I haven't checked how unwieldly the resulting rational equation gets, since there is some uncertainty if the equation has the intended coefficients.
A: Iterating the identity $\cot(\alpha)-\tan(\alpha)=2\cot(2\alpha)$, we get
$$
\cot(\alpha)-\tan(\alpha)-2\tan(2\alpha)-4\tan(4\alpha)-8\tan(8\alpha)=16\cot(16\alpha)\tag{1}
$$
Therefore, using $(1)$, the equation in question becomes
$$
\cot(16\alpha)=\tan(\alpha)\tag{2}
$$
which, due to the periodicity of $\tan$, is equivalent to
$$
n\pi+\frac\pi2-16\alpha=\alpha\tag{3}
$$
That is,
$$
\alpha=\frac\pi{34}(2n+1)\tag{4}
$$
