First slide

Multiple and sub- multiple Angles

Question

$\tan x + \frac{1}{2} \tan \frac{x}{2} + \frac{1}{2^{2}} \tan \frac{x}{2^{2}} + \dots + \frac{1}{2^{n - 1}} \tan \frac{x}{2^{n - 1}}$ is equal to

Moderate

A

$\frac{1}{2^{n}} \cot \frac{x}{2^{n}} - 2 \cot 2 x$
B

$\frac{1}{2^{n - 1}} \cot (\frac{x}{2^{n - 1}}) - 2 \cot 2 x$
C

$\tan \frac{2^{n} - 1}{2^{n - 1}} x$
D

$2 \cot 2 x - \frac{1}{2^{n - 1}} \cot \frac{x}{2^{n - 1}}$

Solution

We can write tan $x = \frac{\tan^{2} x - 1 + 1}{\tan x}$ $= \cot x - 2 \cot 2 x$
$so \begin{array}{l} \tan x + \frac{1}{2} \tan \frac{x}{2} + \frac{1}{2^{2}} \tan \frac{x}{2^{2}} + \dots + \frac{1}{2^{n - 1}} \tan \frac{x}{2^{n - 1}} \\ = \cot x - 2 \cot 2 x + \frac{1}{2} (\cot \frac{x}{2} - 2 \cot x) + \frac{1}{2^{2}} (\cot \frac{x}{2^{2}} - 2 \cot \frac{x}{2}) + \dots + \frac{1}{2^{n - 1}} (\cot \frac{x}{2^{n - 1}} - 2 \cot \frac{x}{2^{n - 2}}) \\ = \frac{1}{2^{n - 1}} \cot (\frac{x}{2^{n - 1}}) - 2 \cot 2 x \end{array}$

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