Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 1 2 Tan ( X 2 ) + 1 4 Tan ( X 4 ) + . . . + 1 2 N Tan ( X 2 N ) = 1 2 N Cot ( X 2 N ) − Cot X for All N ∈ N and 0 < X < π 2 - Mathematics

$\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^n}\tan\left( \frac{x}{2^n} \right) = \frac{1}{2^n}\cot\left( \frac{x}{2^n} \right) - \cot x$ for all n ∈ and  $0 < x < \frac{\pi}{2}$

#### Solution

We need to prove $\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^n}\tan\left( \frac{x}{2^n} \right) = \frac{1}{2^n}\cot\left( \frac{x}{2^n} \right) - \cot x$ for all n ∈ and $0 < x < \frac{\pi}{2}$  using mathematical induction.
For n = 1,
LHS = $\frac{1}{2}\tan\frac{x}{2}$ and

$RHS = \frac{1}{2}\cot\frac{x}{2} - \cot x = \frac{1}{2\tan\frac{x}{2}} - \frac{1}{\tan x}$

$\Rightarrow RHS = \frac{1}{2\tan\frac{x}{2}} - \frac{1}{\frac{2\tan\frac{x}{2}}{1 - \tan^2 \frac{x}{2}}}$

$\Rightarrow RHS = \frac{1}{2\tan\frac{x}{2}} - \frac{1 - \tan^2 \frac{x}{2}}{2 \tan\frac{x}{2}} = \frac{1 - 1 + \tan^2 \frac{x}{2}}{2\tan\frac{x}{2}} = \frac{\tan\frac{x}{2}}{2}$

Therefore, the given relation is true for n = 1.

Now, let the given relation be true for n = k.

We need to prove that the given relation is true for n = k + 1.

$\therefore \frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^k}\tan\left( \frac{x}{2^k} \right) = \frac{1}{2^k}\cot\left( \frac{x}{2^k} \right) - \cot x$
Now, $\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^k}\tan\left( \frac{x}{2^k} \right) + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) = \frac{1}{2^k}\cot\left( \frac{x}{2^k} \right) - \cot x + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right)$
Let:  $L = \frac{1}{2^k}\cot\left( \frac{x}{2^k} \right) - \cot x + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right)$

$\Rightarrow L = \frac{1}{2^k}\cot\left( \frac{x}{2^k} \right) + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x$

$\Rightarrow L = \frac{1}{2^k \tan\left( \frac{x}{2^k} \right)} + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x$

$\Rightarrow L = \frac{1}{2^k \tan2\left( \frac{x}{2^{k + 1}} \right)} + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x$

$\Rightarrow L = \frac{1}{2^k \times \frac{2\tan\left( \frac{x}{2^{k + 1}} \right)}{1 - \tan^2 \left( \frac{x}{2^{k + 1}} \right)}} + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x$

$\Rightarrow L = \frac{1 - \tan^2 \left( \frac{x}{2^{k + 1}} \right)}{2^{k + 1} \tan\left( \frac{x}{2^{k + 1}} \right)} + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) - \cot x$

$\Rightarrow L = \frac{1 - \tan^2 \left( \frac{x}{2^{k + 1}} \right) + \tan^2 \left( \frac{x}{2^{k + 1}} \right)}{2^{k + 1} \tan\left( \frac{x}{2^{k + 1}} \right)} - \cot x = \frac{1}{2^{k + 1}}\cot\left( \frac{x}{2^{k + 1}} \right) - \cot x$

Now,

$\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^k}\tan\left( \frac{x}{2^k} \right) + \frac{1}{2^{k + 1}}\tan\left( \frac{x}{2^{k + 1}} \right) = \frac{1}{2^{k + 1}}\cot\left( \frac{x}{2^{k + 1}} \right) - \cot x$

Thus,

$\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^n}\tan\left( \frac{x}{2^n} \right) = \frac{1}{2^n}\cot\left( \frac{x}{2^n} \right) - \cot x$ for all n ∈ and  $0 < x < \frac{\pi}{2}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 34 | Page 28