Question

Prove the identity $f (x) = \tan x + \frac{1}{2} \tan (\frac{x}{2}) + \frac{1}{2^{2}} \tan (\frac{x}{2^{2}}) + \dots \dots + \frac{1}{2^{n - 1}} \tan (\frac{x}{2^{n - 1}}) = \frac{1}{2^{n - 1}} \cot (\frac{x}{2^{n - 1}}) - 2 \cot 2 x$

Accepted Answer

We have,

$f (x) = \tan x + \frac{1}{2} \tan (\frac{x}{2}) + \frac{1}{2^{2}} \tan (\frac{x}{2^{2}}) + \dots \dots + \frac{1}{2^{n - 1}} \tan (\frac{x}{2^{n - 1}})$

Let

$g (x) = \cos x \cos (\frac{x}{2}) \cos (\frac{x}{2^{2}}) \cos (\frac{x}{2^{3}}) . . . . . . . \cos (\frac{x}{2^{n - 2}}) \cos (\frac{x}{2^{n - 1}})$

$\Rightarrow g (x) = \cos x \cos (\frac{x}{2}) \cos (\frac{x}{2^{2}}) \cos (\frac{x}{2^{3}}) . . . . . . . \cos (\frac{x}{2^{n - 2}}) [2 \cos (\frac{x}{2^{n - 1}}) \sin (\frac{x}{2^{n - 1}})] \frac{1}{2 \sin (\frac{x}{2^{n - 1}})}$

$\Rightarrow g (x) = \cos x \cos (\frac{x}{2}) \cos (\frac{x}{2^{2}}) \cos (\frac{x}{2^{3}}) . . . . . . . [\cos (\frac{x}{2^{n - 2}}) \sin (\frac{x}{2^{n - 2}})] \frac{1}{2 \sin (\frac{x}{2^{n - 1}})}$

$\Rightarrow g (x) = \cos x \cos (\frac{x}{2}) \cos (\frac{x}{2^{2}}) \cos (\frac{x}{2^{3}}) . . . . . . . [2 \cos (\frac{x}{2^{n - 2}}) \sin (\frac{x}{2^{n - 2}})] \frac{1}{2^{2} \sin (\frac{x}{2^{n - 1}})}$

$\Rightarrow g (x) = \cos x \cos (\frac{x}{2}) \cos (\frac{x}{2^{2}}) \cos (\frac{x}{2^{3}}) . . . . . . . [\sin (\frac{x}{2^{n - 3}})] \frac{1}{2^{2} \sin (\frac{x}{2^{n - 1}})}$

$\Rightarrow g (x) = \cos x \cos (\frac{x}{2}) \cos (\frac{x}{2^{2}}) \cos (\frac{x}{2^{3}}) . . . . . . . [\sin (\frac{x}{2^{n - 4}})] \frac{1}{2^{3} \sin (\frac{x}{2^{n - 1}})}$

Proceeding in this manner

$g (x) = \frac{\sin 2 x}{2^{n} \sin (\frac{x}{2^{n - 1}})} . . . . (i)$

Differentiating both sides w.r.t. $x$ , we get

$g^{'} (x) = \frac{1}{2^{n}} [\frac{2 \cos 2 x \sin (\frac{x}{2^{n - 1}}) - (\frac{\sin 2 x}{2^{n - 1}}) \cos (\frac{x}{2^{n - 1}})}{\sin^{2} (\frac{x}{2^{n - 1}})}]$

$\Rightarrow g^{'} (x) = \frac{1}{2^{n}} [\frac{2 \cos 2 x}{\sin (\frac{x}{2^{n - 1}})} - \frac{\sin 2 xcot (\frac{x}{2^{n - 1}})}{2^{n - 1} \sin (\frac{x}{2^{n - 1}})}]$

Hence,

$\frac{g^{'} (x)}{g (x)} = \frac{1}{2^{n}} [\frac{2 \cos 2 x}{\sin (\frac{x}{2^{n - 1}})} - \frac{\sin 2 xcot (\frac{x}{2^{n - 1}})}{2^{n - 1} \sin (\frac{x}{2^{n - 1}})}] \times \frac{2^{n} \sin (\frac{x}{2^{n - 1}})}{\sin 2 x}$

$\frac{g^{'} (x)}{g (x)} = \frac{2 \cos 2 x}{\sin 2 x} - \frac{\sin 2 xcot (\frac{x}{2^{n - 1}})}{2^{n - 1} \sin 2 x}$

$\frac{g^{'} (x)}{g (x)} = 2 \cot 2 x - \frac{1}{2^{n - 1}} \cot (\frac{x}{2^{n - 1}}) . . . . (ii)$

Now,

$g (x) = \cos x \cos (\frac{x}{2}) \cos (\frac{x}{2^{2}}) \cos (\frac{x}{2^{3}}) . . . . . . . \cos (\frac{x}{2^{n - 2}}) \cos (\frac{x}{2^{n - 1}})$

$\Rightarrow \log_{e} g (x) = \log_{e} \cos x + \log_{e} \cos (\frac{x}{2}) + \log_{e} \cos (\frac{x}{2^{2}}) + \log_{e} \cos (\frac{x}{2^{3}}) + . . . . . . . + \log_{e} \cos (\frac{x}{2^{n - 2}}) + \log_{e} \cos (\frac{x}{2^{n - 1}})$

Differentiating both sides w.r.t. $x$ , we get

$\frac{g^{'} (x)}{g (x)} = - \tan x - \frac{1}{2} \tan (\frac{x}{2}) - \frac{1}{2^{2}} \tan (\frac{x}{2^{2}}) - \frac{1}{2^{3}} \tan (\frac{x}{2^{3}}) - . . . . . . . - \frac{1}{2^{n - 1}} \tan (\frac{x}{2^{n - 1}})$

$\Rightarrow \frac{g^{'} (x)}{g (x)} = - f (x)$

$\Rightarrow f (x) = - \frac{g^{'} (x)}{g (x)}$

$\Rightarrow f (x) = 2 \cot 2 x - \frac{1}{2^{n - 1}} \cot (\frac{x}{2^{n - 1}})$

Hence, prove.

Embibe Experts Solutions for Chapter: Trigonometric Ratios, Functions and Identities, Exercise 4: EXERCISE-4

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