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