Question

If $ \cos \frac{x}{2}.\cos \frac{x}{{{2}^{2}}}...\cos \frac{x}{{{2}^{n}}}=\frac{\sin x}{{{2}^{n}}\sin \frac{x}{{{2}^{n}}}}, $ then $ \frac{1}{2}\tan \frac{x}{2}+\frac{1}{{{2}^{2}}}\tan \frac{x}{{{2}^{2}}}+......+\frac{1}{{{2}^{n}}}\tan \frac{x}{{{2}^{n}}} $ is

• 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 $

Answer 1

Answer: (B)

$ \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 $