Explanation

(d)
$2^{n - 1}\tan{(2^{n - 1}\alpha)} + 2^{n}\cot{(2^{n}\alpha)}$
$= 2^{n - 1}\left\lbrack \frac{\sin{2^{n - 1}\alpha}}{\cos{2^{n - 1}\alpha}} + 2\frac{\cos{2^{n}\alpha}}{\sin{2^{n}\alpha}} \right\rbrack$
$= 2^{n - 1}\left\lbrack \frac{\begin{array}{r}
\cos{2^{n}\alpha\cos{2^{n - 1}\alpha + \sin{2^{n}\alpha}}} \\
\sin{2^{n - 1}\alpha + \cos{2^{n}\alpha\cos{2^{n - 1}\alpha}}}
\end{array}}{\sin{2^{n}\alpha\cos{2^{n - 1}\alpha}}} \right\rbrack$
$= 2^{n - 1}\left\lbrack \frac{\cos{2^{n - 1}\alpha(1 + \cos{2^{n}\alpha})}}{\sin{2^{n}\alpha\cos{2^{n - 1}\alpha}}} \right\rbrack$
$= 2^{n - 1}\cot{2^{n - 1}\alpha}$
Proceeding in similar way in last, we get
$\tan{\alpha + 2\cot{2\alpha}}$
$= \frac{\sin\alpha}{\cos\alpha} + 2\frac{\cos{2\alpha}}{\sin{2\alpha}}$
$= \frac{\cos{2\alpha\cos{\alpha + \sin{2\alpha\sin{\alpha + \cos{2\alpha\cos\alpha}}}}}}{\sin{2\alpha\cos\alpha}}$
$= \frac{\cos{\alpha(1 + \cos{2\alpha})}}{2\sin{\alpha\cos^{2}\alpha}}$
$= \frac{2\cos^{2}\alpha}{2\sin\alpha}$
$= \frac{\cos\alpha}{\sin\alpha} = \cot\alpha$