A) \[\cot x-\cot \frac{x}{{{2}^{n}}}\]
B) \[\frac{1}{{{2}^{n}}}\cot \left( \frac{x}{{{2}^{n}}} \right)-\cot x\]
C) \[\frac{1}{{{2}^{n}}}\tan \left( \frac{x}{{{2}^{n}}} \right)-\tan x\]
D) \[\frac{1}{2}\cot x-\frac{1}{{{2}^{n}}}\cot \left( \frac{x}{{{2}^{n}}} \right)\]
E) \[\cot \left( \frac{x}{{{2}^{n}}} \right)-\cot x\]
Correct Answer: B
Solution :
\[\because \]\[\cos \frac{x}{2}.\cos \frac{x}{{{2}^{2}}}.....\cos \frac{x}{{{2}^{n}}}=\frac{\sin x}{{{2}^{n}}\sin \frac{x}{{{2}^{n}}}}\] We have, \[\frac{1}{2}\tan \frac{x}{2}=\frac{1}{2}\cot \frac{x}{2}-\cot x\] and\[\frac{1}{{{2}^{2}}}\tan \frac{x}{{{2}^{n}}}=\frac{1}{{{2}^{n}}}\cot \left( \frac{x}{{{2}^{2}}} \right)-\frac{1}{2}\cot \left( \frac{x}{2} \right)\] Similarly, \[\frac{1}{{{2}^{3}}}\tan \left( \frac{x}{{{2}^{3}}} \right)=\frac{1}{{{2}^{3}}}\cot \left( \frac{x}{{{2}^{3}}} \right)-\frac{1}{{{2}^{2}}}\cot ....\left( \frac{x}{{{2}^{2}}} \right)\] \[\frac{1}{{{2}^{n}}}\tan \left( \frac{x}{{{2}^{n}}} \right)=\frac{1}{{{2}^{n}}}\cot \left( \frac{x}{{{2}^{n}}} \right)-\frac{1}{{{2}^{n-1}}}\cot ....\left( \frac{x}{{{2}^{n-1}}} \right)\] On adding all the above results, we get \[\frac{1}{2}\tan \frac{x}{2}+\frac{1}{{{2}^{2}}}\tan \left( \frac{x}{{{2}^{2}}} \right)+....+\frac{1}{{{2}^{n}}}\tan \left( \frac{x}{{{2}^{n}}} \right)\] \[=\frac{1}{{{2}^{n}}}\cot \left( \frac{x}{{{2}^{n}}} \right)-\cot x\]
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