Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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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  \[0 < x < \frac{\pi}{2}\]

 

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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
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