Explanation

(c)
$\tan{\alpha + 2\tan{2\alpha + 4\tan{4\alpha + 8\cot{8\alpha}}}}$
${= tan}{\alpha + 2\tan{2\alpha + 4\left\lbrack \frac{\sin{4\alpha}}{\cos{4\alpha}} + 2\frac{\cos{8\alpha}}{\sin{8\alpha}} \right\rbrack}}$
${= tan}{\alpha + 2\tan{2\alpha + 4\left\lbrack \frac{\cos{4\alpha\cos{8\alpha + \sin{4\alpha\sin{8\alpha + \cos{4\alpha\cos{8\alpha}}}}}}}{\sin{8\alpha\cos{4\alpha}}} \right\rbrack}}$
$= \tan{\alpha + 2\tan{2\alpha + 4\left\lbrack \frac{\cos{4\alpha + \cos{4\alpha\cos{8\alpha}}}}{\sin{8\alpha\cos{4\alpha}}} \right\rbrack}}$
$= \tan{\alpha + 2\tan{2\alpha + 4\left\lbrack \frac{\cos{4\alpha(1 + \cos{8\alpha})}}{\cos{4\alpha\ \sin\ 8\alpha}} \right\rbrack}}$
$= \tan{\alpha + 2\tan{2\alpha + 4\left\lbrack \frac{2\cos^{2}{4\alpha}}{2\sin{4\alpha\cos{4\alpha}}} \right\rbrack}}$
$= \tan{\alpha + 2(\tan{2\alpha + 2\cot{4\alpha)}}}$
$= \tan{\alpha + 2\left\lbrack \frac{\sin{2\alpha}}{\cos{2\alpha}} + 2\frac{\cos{4\alpha}}{\sin{4\alpha}} \right\rbrack}$
$= \tan{\alpha + 2\left\lbrack \frac{\sin{2\alpha\sin{4\alpha + \cos{4\alpha\cos{2\alpha + \cos{4\alpha\cos{2\alpha}}}}}}}{\sin{4\alpha\cos{2\alpha}}} \right\rbrack}$
$= \tan{\alpha + 2\left\lbrack \frac{\cos{2\alpha + \cos{2\alpha\cos{4\alpha}}}}{\sin{4\alpha\cos{2\alpha}}} \right\rbrack}$
$= \tan{\alpha + 2\ \left\lbrack \frac{\cos{2\alpha(1 + \cos{4\alpha)}}}{\sin{4\alpha\cos{2\alpha}}} \right\rbrack}$
$= \frac{\sin\alpha}{\cos\alpha} + \frac{2\cos{2\alpha}}{\sin{2\alpha}}$
$= \frac{\cos{\alpha + \cos{\alpha\cos{2\alpha}}}}{\sin{2\alpha\cos\alpha}}$
$= \frac{1 + \cos{2\alpha}}{\sin{2\alpha}}$
$= \frac{2\cos^{2}\alpha}{2\sin{\alpha\cos\alpha}} = \cot\alpha$