First slide

Multiple and sub- multiple Angles

Question

For $x \in R$ , $\tan x + \frac{1}{2} \tan \frac{x}{2} + \frac{1}{2^{2}} \tan \frac{x}{2^{2}} + \dots + \frac{1}{2^{n - 1}} \tan (\frac{x}{2^{n - 1}})$ is equal to

Moderate

A

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

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

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

none of these

Solution

We have,
$\begin{array}{l} \tan x = \cot x - 2 \cot 2 x \\ \Rightarrow \frac{1}{2} \tan \frac{x}{2} = \frac{1}{2} \cot \frac{x}{2} - \cot x \\ \Rightarrow \frac{1}{2} \tan \frac{x}{2^{2}} = \frac{1}{2} \cot (\frac{x}{2^{2}}) - \cot (\frac{x}{2}) \\ \Rightarrow \frac{1}{2^{2}} \tan (\frac{x}{2^{2}}) = \frac{1}{2^{2}} \cot (\frac{x}{2^{2}}) - \frac{1}{2} \cot (\frac{x}{2}) \end{array}$
Similarly, we have
$\frac{1}{2^{3}} \tan (\frac{x}{2^{3}}) = \frac{1}{2^{3}} \cot (\frac{x}{2^{3}}) - \frac{1}{2^{2}} \cot (\frac{x}{2^{2}})$
$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$
$\frac{1}{2^{n - 1}} \tan (\frac{x}{2^{n - 1}}) = \frac{1}{2^{n - 1}} \cot (\frac{x}{2^{n - 1}}) - \frac{1}{2^{n - 2}} \cot (\frac{x}{2^{n - 2}})$
Adding a II the above results, we get
$\begin{array}{l} \tan x + \frac{1}{2} \tan \frac{x}{2} + \frac{1}{2^{2}} \tan (\frac{x}{2^{2}}) + \dots + \frac{1}{2^{n - 1}} \tan (\frac{x}{2^{n - 1}}) \\ = \frac{1}{2^{n - 1}} \cot (\frac{x}{2^{n - 1}}) - 2 \cot 2 x \end{array}$

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