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91.
The value of
$\tan (α) + 2 \tan (2 α) + 4 \tan (4 α) + . . . + 2^{n - 1} \tan (2^{n - 1} α) + 2^{n} \cot (2^{n} α)$ is

$\cot (2^{n} α)$

$2^{n} \tan (2^{n} α)$

0

$\cot (α)$

$\cot (α)$

$Now, 2^{n} \tan (2^{n} α) + 2^{n} \cot (2^{n} α) = 2^{n - 1} [\frac{\sin (2^{n - 1} α)}{\cos (2^{n - 1} α)} + 2 \frac{\cos (2^{n} α)}{\sin (2^{n} α)}] = 2^{n - 1} [\frac{\cos (2^{n} α) \cos (2^{n - 1} α) + \sin (2^{n} α) \sin (2^{n - 1} α) + \cos (2^{n} α) \cos (2^{n - 1} α)}{\sin (2^{n} α) \cos (2^{n - 1} α)}] = 2^{n - 1} [\frac{\cos (2^{n - 1} α) (1 + \cos (2^{n} α))}{\sin (2^{n} α) \cos (2^{n - 1} α)}] = 2^{n - 1} \cot (2^{n - 1} α)$

Proceeding in similar way in last, we get

$\tan (α) + 2 \cot (2 α)$

$= \frac{\sin (α)}{\cos (α)} + 2 \frac{\cos (2 α)}{\sin (2 α)} = \frac{\cos (2 α) \cos (α) + \sin (2 α) \sin (α) + \cos (2 α) \cos (α)}{\sin (2 α) \cos (α)} = \frac{\cos (α) (1 + \cos (2 α))}{2 \sin (α) \cos^{2} (α)} = \frac{2 \cos (α)}{2 \sin (α)} = \frac{\cos (α)}{\sin (α)} = \cot (α)$